Use a table of values to graph the following functions. State the domain and range.���y = x2 + 2x + 5

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The graph opens upwards, so the vertex is a minimum located at (í1, 4). So the range is all real numbers greater than or equal to the minimum value of 4, or R = {y | y ����`�

x y �í3 8 í2 5 í1 4 0 5 1 8 2 13

���y = 2x2 í 3x + 1

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The range is all real numbers greater than or equal to the minimum value.

R = {y | y �� í0.125}

x y í2 15 í1 6 0 1 1 0 2 3 3 10

Consider y = x2 í 7x + 6.���Determine whether the function has a maximum or minimum value.

62/87,21���For y = x2 ± 7x + 6, a = 1, b = ±7, and c = 6. Because a is positive, the graph opens upward, so the function has a minimum value.

���State the maximum or minimum value.

62/87,21���Method 1: Find the vertex of the parabola. First, find the x-coordinate of the vertex. For the function y = x2 ± 7x + 6, use a = 1, b = ±7, and c = 6. �

� Substitute 3.5 for x in the function to find the value of the y -FRRUGLQDWH� �

� So, the minimum value is ±6.25. Method 2: Use a graphing calculator to graph the related function f (x) = x2 ± 7x + 6. Use the minimum option on the 2nd [CALC] menu to determine the minimum value.

Therefore, the minimum value is ±6.25.

���What are the domain and range?

62/87,21���Any number can be substituted for x in the function. So, tKH�GRPDLQ�LV�'� �^DOO�UHDO�QXPEHUV`�� The minimum of the function is í6.25. Therefore, R = {y | y �� í6.25} because the range is all real numbers that are JUHDWHU�WKDQ�RU�HTXDO�WR�WKH�PLQLPXP�YDOXH��

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

���x2 + 7x + 10 = 0

62/87,21���Graph the related function f (x) = x2 + 7x + 10.

The x-intercepts of the graph appear to be at ±5 and ±2, so the solutions are ±5 and ±2.Check:

and

���x2 í 5 = í3x

62/87,21���Write the equation in standard form.

Graph the related function f (x) = x2 + 3x í 5.

The x-intercepts are located between ±5 and ±4 and between 1 and 2. Make a table using an increment of 0.1 for the x-values located between ±5 and ±4 and between 1 and 2.

�

For each table, the function value that is closest to zero when the sign changes is 0.04. Thus, the roots are approximately ±4.2 and 1.2.

x ±4.5 ±4.4 ±4.3 ±4.2 ±4.1 y 1.75 1.16 0.59 0.04 ±0.49

x 1.1 1.2 1.3 1.4 1.5 y ±0.49 0.04 0.59 1.16 1.75

Describe how the graph of each function is related to the graph of f (x) = x2.���g(x) = x2 í 5

62/87,21���

The graph of f (x) = x2 + c represents a vertical translation of the parent graph. The value of c is ±5, and ±5 < 0. If c < 0, the graph of f (x) = x2 is translated �XQLWV�GRZQ��7KHUHIRUH��WKH�JUDSK�RI�y = x2 ± 5 is a translation of the graph

of y = x2 shifted down 5 units.

���g(x) = í3x2

62/87,21���

�7KH�JUDSK�RI�f (x) = ax2 expands or compresses the parent graph vertically. Since |a| > 1, the graph is vertically expanded. Since a < 1, the graph is reflected across the x-axis. So, the graph of y = í3x2 is the graph of y = x2 reflected across the x-axis and vertically expanded. �

����

62/87,21���

The function can be written f (x) = ax2 + c, where a = �DQG�c = 4. Since 4 > 0 and 0 < ������WKH�JUDSK�RI�y =

x2 + 4 is the graph of y = x2 vertically compressed and shifted up 4 units.

����08/7,3/(�&+2,&(� Which is an equation for the function shown in the graph?

$� y = í3x2 %� y = 3x2 + 1 &� y = x2 + 2 '� y = í3x2 + 2

62/87,21���The parabola opens downward, so the a in the equation must be negative. Therefore, choices B and C can be eliminated. The y-intercept is at the point (0, c), or in this case (0, 2). Therefore c = 2, so choice A can be eliminated.So, the correct choice is choice D.

Solve each equation by completing the square.����x2 + 2x + 5 = 0

62/87,21���

� No real number has a negative square. So, this equation has no real solutions.

����x2 í x í 6 = 0

62/87,21���

� The solutions are 3 and ±2.

����2x2 í 36 = í6x

62/87,21���Rewrite the equation in standard form.

� The solutions are 3 and ±6.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.����x2 í x í 30 = 0

62/87,21���For this equation, a = 1, b = ±1, and c = ±30.

� The solutions are 6 and ±5.

����x2 í 10x = í15

62/87,21���Rewrite the equation in standard form. �

� For this equation, a = 1, b = ±10, and c = 15.

� The solutions are 1.8 and 8.2.

����2x2 + x ± 15 = 0

62/87,21���For this equation, a = 2, b = 1, and c = ±15.

� The solutions are 2.5 and ±3.

����%$6(%$//��Elias hits a baseball into the air. The equation h = ±16t2 + 60t +3 models the height h in feet of the ball after t seconds. How long is the ball in the air?

62/87,21���

� The ball is in the air for about 3.8 seconds.

����Graph {(í2, 4), (í1, 1), (0, 0), (1, 1), (2, 4)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

62/87,21���

The points are reflected across a line of symmetry. They represent a quadratic function.

����Look for a pattern in the table to determine which kind of model best describes the data.

62/87,21���

� Calculate the first differences.

6LQFH�WKH�ILUVW�GLIIHUHQFHV�DUH�HTXDO��D�OLQHDU�IXQFWLRQ�PRGHOV�WKH�GDWD��Find the slope and y -LQWHUFHSW��

� Use the slope and one point from the table to find the y -LQWHUFHSW��

The data describes the linear function f (x) = 2x + 1.

����CAR CLUB The table shows the number of car club members for four consecutive years after it began. �

� a. Determine which model best represents the data. b. Write a function that models the data. c. Predict the number of car club members after 6 years.

62/87,21���a. First compare the first differences in the number of members:

� Since they're not equal, compare the second differences:

� Since the second differences are not equal, look for a common ratio:

� 7KHUH�LV�D�FRPPRQ�UDWLR�RI����VR�WKH�EHVW�PRGHO�ZRXOG�EH�DQ�H[SRQHQWLDO�IXQFWLRQ�WR�ILW�WKH�GDWD�� � b. A function that fits the data should be an exponential function with a ratio of 2. We are looking for a function of the form . Use a point from the data given to determine the value of a.

� � The function that best fits the data is:

� c. To find the number of car club members after 6 years, use the function found in part b and plug in 6 for x.

7KHUH�ZLOO�EH�����PHPEHUV�DIWHU���\HDUV���

Graph each function.����f (x) = |x í 1|

62/87,21���Make a table of values. �

�

x f (x) ±3 4 ±2 3 ±1 2 0 1 1 0 2 1 3 2 4 3

����f (x) = í|2x|

62/87,21���The negative in front of the absolute value makes the graph point down. Make a table of values.�

�

x f (x) ±2 ±4 ±1 ±2 0 0 1 ±2 2 ±4

����f (x) =

62/87,21���Make a table of values. �

�

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

����f (x) =

62/87,21���This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. �

�

x f (x) ±2 ±5 ±1 ±3 0 ±1 1 1 2 ±1 3 0 4 1

����Determine the domain and range of the function graphed below.

62/87,21���The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = �í 2, so the domain is all real numbers and the range is {y | y �í 2}.

eSolutions Manual - Powered by Cognero Page 1

Practice Test - Chapter 9

Use a table of values to graph the following functions. State the domain and range.���y = x2 + 2x + 5

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The graph opens upwards, so the vertex is a minimum located at (í1, 4). So the range is all real numbers greater than or equal to the minimum value of 4, or R = {y | y ����`�

x y �í3 8 í2 5 í1 4 0 5 1 8 2 13

���y = 2x2 í 3x + 1

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The range is all real numbers greater than or equal to the minimum value.

R = {y | y �� í0.125}

x y í2 15 í1 6 0 1 1 0 2 3 3 10

Consider y = x2 í 7x + 6.���Determine whether the function has a maximum or minimum value.

62/87,21���For y = x2 ± 7x + 6, a = 1, b = ±7, and c = 6. Because a is positive, the graph opens upward, so the function has a minimum value.

���State the maximum or minimum value.

62/87,21���Method 1: Find the vertex of the parabola. First, find the x-coordinate of the vertex. For the function y = x2 ± 7x + 6, use a = 1, b = ±7, and c = 6. �

� Substitute 3.5 for x in the function to find the value of the y -FRRUGLQDWH� �

� So, the minimum value is ±6.25. Method 2: Use a graphing calculator to graph the related function f (x) = x2 ± 7x + 6. Use the minimum option on the 2nd [CALC] menu to determine the minimum value.

Therefore, the minimum value is ±6.25.

���What are the domain and range?

62/87,21���Any number can be substituted for x in the function. So, tKH�GRPDLQ�LV�'� �^DOO�UHDO�QXPEHUV`�� The minimum of the function is í6.25. Therefore, R = {y | y �� í6.25} because the range is all real numbers that are JUHDWHU�WKDQ�RU�HTXDO�WR�WKH�PLQLPXP�YDOXH��

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

���x2 + 7x + 10 = 0

62/87,21���Graph the related function f (x) = x2 + 7x + 10.

The x-intercepts of the graph appear to be at ±5 and ±2, so the solutions are ±5 and ±2.Check:

and

���x2 í 5 = í3x

62/87,21���Write the equation in standard form.

Graph the related function f (x) = x2 + 3x í 5.

The x-intercepts are located between ±5 and ±4 and between 1 and 2. Make a table using an increment of 0.1 for the x-values located between ±5 and ±4 and between 1 and 2.

�

For each table, the function value that is closest to zero when the sign changes is 0.04. Thus, the roots are approximately ±4.2 and 1.2.

x ±4.5 ±4.4 ±4.3 ±4.2 ±4.1 y 1.75 1.16 0.59 0.04 ±0.49

x 1.1 1.2 1.3 1.4 1.5 y ±0.49 0.04 0.59 1.16 1.75

Describe how the graph of each function is related to the graph of f (x) = x2.���g(x) = x2 í 5

62/87,21���

The graph of f (x) = x2 + c represents a vertical translation of the parent graph. The value of c is ±5, and ±5 < 0. If c < 0, the graph of f (x) = x2 is translated �XQLWV�GRZQ��7KHUHIRUH��WKH�JUDSK�RI�y = x2 ± 5 is a translation of the graph

of y = x2 shifted down 5 units.

���g(x) = í3x2

62/87,21���

�7KH�JUDSK�RI�f (x) = ax2 expands or compresses the parent graph vertically. Since |a| > 1, the graph is vertically expanded. Since a < 1, the graph is reflected across the x-axis. So, the graph of y = í3x2 is the graph of y = x2 reflected across the x-axis and vertically expanded. �

����

62/87,21���

The function can be written f (x) = ax2 + c, where a = �DQG�c = 4. Since 4 > 0 and 0 < ������WKH�JUDSK�RI�y =

x2 + 4 is the graph of y = x2 vertically compressed and shifted up 4 units.

����08/7,3/(�&+2,&(� Which is an equation for the function shown in the graph?

$� y = í3x2 %� y = 3x2 + 1 &� y = x2 + 2 '� y = í3x2 + 2

62/87,21���The parabola opens downward, so the a in the equation must be negative. Therefore, choices B and C can be eliminated. The y-intercept is at the point (0, c), or in this case (0, 2). Therefore c = 2, so choice A can be eliminated.So, the correct choice is choice D.

Solve each equation by completing the square.����x2 + 2x + 5 = 0

62/87,21���

� No real number has a negative square. So, this equation has no real solutions.

����x2 í x í 6 = 0

62/87,21���

� The solutions are 3 and ±2.

����2x2 í 36 = í6x

62/87,21���Rewrite the equation in standard form.

� The solutions are 3 and ±6.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.����x2 í x í 30 = 0

62/87,21���For this equation, a = 1, b = ±1, and c = ±30.

� The solutions are 6 and ±5.

����x2 í 10x = í15

62/87,21���Rewrite the equation in standard form. �

� For this equation, a = 1, b = ±10, and c = 15.

� The solutions are 1.8 and 8.2.

����2x2 + x ± 15 = 0

62/87,21���For this equation, a = 2, b = 1, and c = ±15.

� The solutions are 2.5 and ±3.

����%$6(%$//��Elias hits a baseball into the air. The equation h = ±16t2 + 60t +3 models the height h in feet of the ball after t seconds. How long is the ball in the air?

62/87,21���

� The ball is in the air for about 3.8 seconds.

����Graph {(í2, 4), (í1, 1), (0, 0), (1, 1), (2, 4)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

62/87,21���

The points are reflected across a line of symmetry. They represent a quadratic function.

����Look for a pattern in the table to determine which kind of model best describes the data.

62/87,21���

� Calculate the first differences.

6LQFH�WKH�ILUVW�GLIIHUHQFHV�DUH�HTXDO��D�OLQHDU�IXQFWLRQ�PRGHOV�WKH�GDWD��Find the slope and y -LQWHUFHSW��

� Use the slope and one point from the table to find the y -LQWHUFHSW��

The data describes the linear function f (x) = 2x + 1.

����CAR CLUB The table shows the number of car club members for four consecutive years after it began. �

� a. Determine which model best represents the data. b. Write a function that models the data. c. Predict the number of car club members after 6 years.

62/87,21���a. First compare the first differences in the number of members:

� Since they're not equal, compare the second differences:

� Since the second differences are not equal, look for a common ratio:

� 7KHUH�LV�D�FRPPRQ�UDWLR�RI����VR�WKH�EHVW�PRGHO�ZRXOG�EH�DQ�H[SRQHQWLDO�IXQFWLRQ�WR�ILW�WKH�GDWD�� � b. A function that fits the data should be an exponential function with a ratio of 2. We are looking for a function of the form . Use a point from the data given to determine the value of a.

� � The function that best fits the data is:

� c. To find the number of car club members after 6 years, use the function found in part b and plug in 6 for x.

7KHUH�ZLOO�EH�����PHPEHUV�DIWHU���\HDUV���

Graph each function.����f (x) = |x í 1|

62/87,21���Make a table of values. �

�

x f (x) ±3 4 ±2 3 ±1 2 0 1 1 0 2 1 3 2 4 3

����f (x) = í|2x|

62/87,21���The negative in front of the absolute value makes the graph point down. Make a table of values.�

�

x f (x) ±2 ±4 ±1 ±2 0 0 1 ±2 2 ±4

����f (x) =

62/87,21���Make a table of values. �

�

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

����f (x) =

62/87,21���This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. �

�

x f (x) ±2 ±5 ±1 ±3 0 ±1 1 1 2 ±1 3 0 4 1

����Determine the domain and range of the function graphed below.

62/87,21���The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = �í 2, so the domain is all real numbers and the range is {y | y �í 2}.

eSolutions Manual - Powered by Cognero Page 2

Practice Test - Chapter 9

Use a table of values to graph the following functions. State the domain and range.���y = x2 + 2x + 5

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The graph opens upwards, so the vertex is a minimum located at (í1, 4). So the range is all real numbers greater than or equal to the minimum value of 4, or R = {y | y ����`�

x y �í3 8 í2 5 í1 4 0 5 1 8 2 13

���y = 2x2 í 3x + 1

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The range is all real numbers greater than or equal to the minimum value.

R = {y | y �� í0.125}

x y í2 15 í1 6 0 1 1 0 2 3 3 10

Consider y = x2 í 7x + 6.���Determine whether the function has a maximum or minimum value.

62/87,21���For y = x2 ± 7x + 6, a = 1, b = ±7, and c = 6. Because a is positive, the graph opens upward, so the function has a minimum value.

���State the maximum or minimum value.

62/87,21���Method 1: Find the vertex of the parabola. First, find the x-coordinate of the vertex. For the function y = x2 ± 7x + 6, use a = 1, b = ±7, and c = 6. �

� Substitute 3.5 for x in the function to find the value of the y -FRRUGLQDWH� �

� So, the minimum value is ±6.25. Method 2: Use a graphing calculator to graph the related function f (x) = x2 ± 7x + 6. Use the minimum option on the 2nd [CALC] menu to determine the minimum value.

Therefore, the minimum value is ±6.25.

���What are the domain and range?

62/87,21���Any number can be substituted for x in the function. So, tKH�GRPDLQ�LV�'� �^DOO�UHDO�QXPEHUV`�� The minimum of the function is í6.25. Therefore, R = {y | y �� í6.25} because the range is all real numbers that are JUHDWHU�WKDQ�RU�HTXDO�WR�WKH�PLQLPXP�YDOXH��

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

���x2 + 7x + 10 = 0

62/87,21���Graph the related function f (x) = x2 + 7x + 10.

The x-intercepts of the graph appear to be at ±5 and ±2, so the solutions are ±5 and ±2.Check:

and

���x2 í 5 = í3x

62/87,21���Write the equation in standard form.

Graph the related function f (x) = x2 + 3x í 5.

The x-intercepts are located between ±5 and ±4 and between 1 and 2. Make a table using an increment of 0.1 for the x-values located between ±5 and ±4 and between 1 and 2.

�

For each table, the function value that is closest to zero when the sign changes is 0.04. Thus, the roots are approximately ±4.2 and 1.2.

x ±4.5 ±4.4 ±4.3 ±4.2 ±4.1 y 1.75 1.16 0.59 0.04 ±0.49

x 1.1 1.2 1.3 1.4 1.5 y ±0.49 0.04 0.59 1.16 1.75

Describe how the graph of each function is related to the graph of f (x) = x2.���g(x) = x2 í 5

62/87,21���

The graph of f (x) = x2 + c represents a vertical translation of the parent graph. The value of c is ±5, and ±5 < 0. If c < 0, the graph of f (x) = x2 is translated �XQLWV�GRZQ��7KHUHIRUH��WKH�JUDSK�RI�y = x2 ± 5 is a translation of the graph

of y = x2 shifted down 5 units.

���g(x) = í3x2

62/87,21���

�7KH�JUDSK�RI�f (x) = ax2 expands or compresses the parent graph vertically. Since |a| > 1, the graph is vertically expanded. Since a < 1, the graph is reflected across the x-axis. So, the graph of y = í3x2 is the graph of y = x2 reflected across the x-axis and vertically expanded. �

����

62/87,21���

The function can be written f (x) = ax2 + c, where a = �DQG�c = 4. Since 4 > 0 and 0 < ������WKH�JUDSK�RI�y =

x2 + 4 is the graph of y = x2 vertically compressed and shifted up 4 units.

����08/7,3/(�&+2,&(� Which is an equation for the function shown in the graph?

$� y = í3x2 %� y = 3x2 + 1 &� y = x2 + 2 '� y = í3x2 + 2

62/87,21���The parabola opens downward, so the a in the equation must be negative. Therefore, choices B and C can be eliminated. The y-intercept is at the point (0, c), or in this case (0, 2). Therefore c = 2, so choice A can be eliminated.So, the correct choice is choice D.

Solve each equation by completing the square.����x2 + 2x + 5 = 0

62/87,21���

� No real number has a negative square. So, this equation has no real solutions.

����x2 í x í 6 = 0

62/87,21���

� The solutions are 3 and ±2.

����2x2 í 36 = í6x

62/87,21���Rewrite the equation in standard form.

� The solutions are 3 and ±6.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.����x2 í x í 30 = 0

62/87,21���For this equation, a = 1, b = ±1, and c = ±30.

� The solutions are 6 and ±5.

����x2 í 10x = í15

62/87,21���Rewrite the equation in standard form. �

� For this equation, a = 1, b = ±10, and c = 15.

� The solutions are 1.8 and 8.2.

����2x2 + x ± 15 = 0

62/87,21���For this equation, a = 2, b = 1, and c = ±15.

� The solutions are 2.5 and ±3.

����%$6(%$//��Elias hits a baseball into the air. The equation h = ±16t2 + 60t +3 models the height h in feet of the ball after t seconds. How long is the ball in the air?

62/87,21���

� The ball is in the air for about 3.8 seconds.

����Graph {(í2, 4), (í1, 1), (0, 0), (1, 1), (2, 4)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

62/87,21���

The points are reflected across a line of symmetry. They represent a quadratic function.

����Look for a pattern in the table to determine which kind of model best describes the data.

62/87,21���

� Calculate the first differences.

6LQFH�WKH�ILUVW�GLIIHUHQFHV�DUH�HTXDO��D�OLQHDU�IXQFWLRQ�PRGHOV�WKH�GDWD��Find the slope and y -LQWHUFHSW��

� Use the slope and one point from the table to find the y -LQWHUFHSW��

The data describes the linear function f (x) = 2x + 1.

����CAR CLUB The table shows the number of car club members for four consecutive years after it began. �

� a. Determine which model best represents the data. b. Write a function that models the data. c. Predict the number of car club members after 6 years.

62/87,21���a. First compare the first differences in the number of members:

� Since they're not equal, compare the second differences:

� Since the second differences are not equal, look for a common ratio:

� 7KHUH�LV�D�FRPPRQ�UDWLR�RI����VR�WKH�EHVW�PRGHO�ZRXOG�EH�DQ�H[SRQHQWLDO�IXQFWLRQ�WR�ILW�WKH�GDWD�� � b. A function that fits the data should be an exponential function with a ratio of 2. We are looking for a function of the form . Use a point from the data given to determine the value of a.

� � The function that best fits the data is:

� c. To find the number of car club members after 6 years, use the function found in part b and plug in 6 for x.

7KHUH�ZLOO�EH�����PHPEHUV�DIWHU���\HDUV���

Graph each function.����f (x) = |x í 1|

62/87,21���Make a table of values. �

�

x f (x) ±3 4 ±2 3 ±1 2 0 1 1 0 2 1 3 2 4 3

����f (x) = í|2x|

62/87,21���The negative in front of the absolute value makes the graph point down. Make a table of values.�

�

x f (x) ±2 ±4 ±1 ±2 0 0 1 ±2 2 ±4

����f (x) =

62/87,21���Make a table of values. �

�

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

����f (x) =

62/87,21���This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. �

�

x f (x) ±2 ±5 ±1 ±3 0 ±1 1 1 2 ±1 3 0 4 1

����Determine the domain and range of the function graphed below.

62/87,21���The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = �í 2, so the domain is all real numbers and the range is {y | y �í 2}.

eSolutions Manual - Powered by Cognero Page 3

Practice Test - Chapter 9

Use a table of values to graph the following functions. State the domain and range.���y = x2 + 2x + 5

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The graph opens upwards, so the vertex is a minimum located at (í1, 4). So the range is all real numbers greater than or equal to the minimum value of 4, or R = {y | y ����`�

x y �í3 8 í2 5 í1 4 0 5 1 8 2 13

���y = 2x2 í 3x + 1

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The range is all real numbers greater than or equal to the minimum value.

R = {y | y �� í0.125}

x y í2 15 í1 6 0 1 1 0 2 3 3 10

Consider y = x2 í 7x + 6.���Determine whether the function has a maximum or minimum value.

62/87,21���For y = x2 ± 7x + 6, a = 1, b = ±7, and c = 6. Because a is positive, the graph opens upward, so the function has a minimum value.

���State the maximum or minimum value.

62/87,21���Method 1: Find the vertex of the parabola. First, find the x-coordinate of the vertex. For the function y = x2 ± 7x + 6, use a = 1, b = ±7, and c = 6. �

� Substitute 3.5 for x in the function to find the value of the y -FRRUGLQDWH� �

� So, the minimum value is ±6.25. Method 2: Use a graphing calculator to graph the related function f (x) = x2 ± 7x + 6. Use the minimum option on the 2nd [CALC] menu to determine the minimum value.

Therefore, the minimum value is ±6.25.

���What are the domain and range?

62/87,21���Any number can be substituted for x in the function. So, tKH�GRPDLQ�LV�'� �^DOO�UHDO�QXPEHUV`�� The minimum of the function is í6.25. Therefore, R = {y | y �� í6.25} because the range is all real numbers that are JUHDWHU�WKDQ�RU�HTXDO�WR�WKH�PLQLPXP�YDOXH��

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

���x2 + 7x + 10 = 0

62/87,21���Graph the related function f (x) = x2 + 7x + 10.

The x-intercepts of the graph appear to be at ±5 and ±2, so the solutions are ±5 and ±2.Check:

and

���x2 í 5 = í3x

62/87,21���Write the equation in standard form.

Graph the related function f (x) = x2 + 3x í 5.

The x-intercepts are located between ±5 and ±4 and between 1 and 2. Make a table using an increment of 0.1 for the x-values located between ±5 and ±4 and between 1 and 2.

�

For each table, the function value that is closest to zero when the sign changes is 0.04. Thus, the roots are approximately ±4.2 and 1.2.

x ±4.5 ±4.4 ±4.3 ±4.2 ±4.1 y 1.75 1.16 0.59 0.04 ±0.49

x 1.1 1.2 1.3 1.4 1.5 y ±0.49 0.04 0.59 1.16 1.75

Describe how the graph of each function is related to the graph of f (x) = x2.���g(x) = x2 í 5

62/87,21���

The graph of f (x) = x2 + c represents a vertical translation of the parent graph. The value of c is ±5, and ±5 < 0. If c < 0, the graph of f (x) = x2 is translated �XQLWV�GRZQ��7KHUHIRUH��WKH�JUDSK�RI�y = x2 ± 5 is a translation of the graph

of y = x2 shifted down 5 units.

���g(x) = í3x2

62/87,21���

�7KH�JUDSK�RI�f (x) = ax2 expands or compresses the parent graph vertically. Since |a| > 1, the graph is vertically expanded. Since a < 1, the graph is reflected across the x-axis. So, the graph of y = í3x2 is the graph of y = x2 reflected across the x-axis and vertically expanded. �

����

62/87,21���

The function can be written f (x) = ax2 + c, where a = �DQG�c = 4. Since 4 > 0 and 0 < ������WKH�JUDSK�RI�y =

x2 + 4 is the graph of y = x2 vertically compressed and shifted up 4 units.

����08/7,3/(�&+2,&(� Which is an equation for the function shown in the graph?

$� y = í3x2 %� y = 3x2 + 1 &� y = x2 + 2 '� y = í3x2 + 2

62/87,21���The parabola opens downward, so the a in the equation must be negative. Therefore, choices B and C can be eliminated. The y-intercept is at the point (0, c), or in this case (0, 2). Therefore c = 2, so choice A can be eliminated.So, the correct choice is choice D.

Solve each equation by completing the square.����x2 + 2x + 5 = 0

62/87,21���

� No real number has a negative square. So, this equation has no real solutions.

����x2 í x í 6 = 0

62/87,21���

� The solutions are 3 and ±2.

����2x2 í 36 = í6x

62/87,21���Rewrite the equation in standard form.

� The solutions are 3 and ±6.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.����x2 í x í 30 = 0

62/87,21���For this equation, a = 1, b = ±1, and c = ±30.

� The solutions are 6 and ±5.

����x2 í 10x = í15

62/87,21���Rewrite the equation in standard form. �

� For this equation, a = 1, b = ±10, and c = 15.

� The solutions are 1.8 and 8.2.

����2x2 + x ± 15 = 0

62/87,21���For this equation, a = 2, b = 1, and c = ±15.

� The solutions are 2.5 and ±3.

����%$6(%$//��Elias hits a baseball into the air. The equation h = ±16t2 + 60t +3 models the height h in feet of the ball after t seconds. How long is the ball in the air?

62/87,21���

� The ball is in the air for about 3.8 seconds.

����Graph {(í2, 4), (í1, 1), (0, 0), (1, 1), (2, 4)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

62/87,21���

The points are reflected across a line of symmetry. They represent a quadratic function.

����Look for a pattern in the table to determine which kind of model best describes the data.

62/87,21���

� Calculate the first differences.

6LQFH�WKH�ILUVW�GLIIHUHQFHV�DUH�HTXDO��D�OLQHDU�IXQFWLRQ�PRGHOV�WKH�GDWD��Find the slope and y -LQWHUFHSW��

� Use the slope and one point from the table to find the y -LQWHUFHSW��

The data describes the linear function f (x) = 2x + 1.

����CAR CLUB The table shows the number of car club members for four consecutive years after it began. �

� a. Determine which model best represents the data. b. Write a function that models the data. c. Predict the number of car club members after 6 years.

62/87,21���a. First compare the first differences in the number of members:

� Since they're not equal, compare the second differences:

� Since the second differences are not equal, look for a common ratio:

� 7KHUH�LV�D�FRPPRQ�UDWLR�RI����VR�WKH�EHVW�PRGHO�ZRXOG�EH�DQ�H[SRQHQWLDO�IXQFWLRQ�WR�ILW�WKH�GDWD�� � b. A function that fits the data should be an exponential function with a ratio of 2. We are looking for a function of the form . Use a point from the data given to determine the value of a.

� � The function that best fits the data is:

� c. To find the number of car club members after 6 years, use the function found in part b and plug in 6 for x.

7KHUH�ZLOO�EH�����PHPEHUV�DIWHU���\HDUV���

Graph each function.����f (x) = |x í 1|

62/87,21���Make a table of values. �

�

x f (x) ±3 4 ±2 3 ±1 2 0 1 1 0 2 1 3 2 4 3

����f (x) = í|2x|

62/87,21���The negative in front of the absolute value makes the graph point down. Make a table of values.�

�

x f (x) ±2 ±4 ±1 ±2 0 0 1 ±2 2 ±4

����f (x) =

62/87,21���Make a table of values. �

�

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

����f (x) =

62/87,21���This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. �

�

x f (x) ±2 ±5 ±1 ±3 0 ±1 1 1 2 ±1 3 0 4 1

����Determine the domain and range of the function graphed below.

62/87,21���The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = �í 2, so the domain is all real numbers and the range is {y | y �í 2}.

eSolutions Manual - Powered by Cognero Page 4

Practice Test - Chapter 9

Use a table of values to graph the following functions. State the domain and range.���y = x2 + 2x + 5

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The graph opens upwards, so the vertex is a minimum located at (í1, 4). So the range is all real numbers greater than or equal to the minimum value of 4, or R = {y | y ����`�

x y �í3 8 í2 5 í1 4 0 5 1 8 2 13

���y = 2x2 í 3x + 1

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The range is all real numbers greater than or equal to the minimum value.

R = {y | y �� í0.125}

x y í2 15 í1 6 0 1 1 0 2 3 3 10

Consider y = x2 í 7x + 6.���Determine whether the function has a maximum or minimum value.

62/87,21���For y = x2 ± 7x + 6, a = 1, b = ±7, and c = 6. Because a is positive, the graph opens upward, so the function has a minimum value.

���State the maximum or minimum value.

62/87,21���Method 1: Find the vertex of the parabola. First, find the x-coordinate of the vertex. For the function y = x2 ± 7x + 6, use a = 1, b = ±7, and c = 6. �

� Substitute 3.5 for x in the function to find the value of the y -FRRUGLQDWH� �

� So, the minimum value is ±6.25. Method 2: Use a graphing calculator to graph the related function f (x) = x2 ± 7x + 6. Use the minimum option on the 2nd [CALC] menu to determine the minimum value.

Therefore, the minimum value is ±6.25.

���What are the domain and range?

62/87,21���Any number can be substituted for x in the function. So, tKH�GRPDLQ�LV�'� �^DOO�UHDO�QXPEHUV`�� The minimum of the function is í6.25. Therefore, R = {y | y �� í6.25} because the range is all real numbers that are JUHDWHU�WKDQ�RU�HTXDO�WR�WKH�PLQLPXP�YDOXH��

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

���x2 + 7x + 10 = 0

62/87,21���Graph the related function f (x) = x2 + 7x + 10.

The x-intercepts of the graph appear to be at ±5 and ±2, so the solutions are ±5 and ±2.Check:

and

���x2 í 5 = í3x

62/87,21���Write the equation in standard form.

Graph the related function f (x) = x2 + 3x í 5.

The x-intercepts are located between ±5 and ±4 and between 1 and 2. Make a table using an increment of 0.1 for the x-values located between ±5 and ±4 and between 1 and 2.

�

For each table, the function value that is closest to zero when the sign changes is 0.04. Thus, the roots are approximately ±4.2 and 1.2.

x ±4.5 ±4.4 ±4.3 ±4.2 ±4.1 y 1.75 1.16 0.59 0.04 ±0.49

x 1.1 1.2 1.3 1.4 1.5 y ±0.49 0.04 0.59 1.16 1.75

Describe how the graph of each function is related to the graph of f (x) = x2.���g(x) = x2 í 5

62/87,21���

The graph of f (x) = x2 + c represents a vertical translation of the parent graph. The value of c is ±5, and ±5 < 0. If c < 0, the graph of f (x) = x2 is translated �XQLWV�GRZQ��7KHUHIRUH��WKH�JUDSK�RI�y = x2 ± 5 is a translation of the graph

of y = x2 shifted down 5 units.

���g(x) = í3x2

62/87,21���

�7KH�JUDSK�RI�f (x) = ax2 expands or compresses the parent graph vertically. Since |a| > 1, the graph is vertically expanded. Since a < 1, the graph is reflected across the x-axis. So, the graph of y = í3x2 is the graph of y = x2 reflected across the x-axis and vertically expanded. �

����

62/87,21���

The function can be written f (x) = ax2 + c, where a = �DQG�c = 4. Since 4 > 0 and 0 < ������WKH�JUDSK�RI�y =

x2 + 4 is the graph of y = x2 vertically compressed and shifted up 4 units.

����08/7,3/(�&+2,&(� Which is an equation for the function shown in the graph?

$� y = í3x2 %� y = 3x2 + 1 &� y = x2 + 2 '� y = í3x2 + 2

62/87,21���The parabola opens downward, so the a in the equation must be negative. Therefore, choices B and C can be eliminated. The y-intercept is at the point (0, c), or in this case (0, 2). Therefore c = 2, so choice A can be eliminated.So, the correct choice is choice D.

Solve each equation by completing the square.����x2 + 2x + 5 = 0

62/87,21���

� No real number has a negative square. So, this equation has no real solutions.

����x2 í x í 6 = 0

62/87,21���

� The solutions are 3 and ±2.

����2x2 í 36 = í6x

62/87,21���Rewrite the equation in standard form.

� The solutions are 3 and ±6.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.����x2 í x í 30 = 0

62/87,21���For this equation, a = 1, b = ±1, and c = ±30.

� The solutions are 6 and ±5.

����x2 í 10x = í15

62/87,21���Rewrite the equation in standard form. �

� For this equation, a = 1, b = ±10, and c = 15.

� The solutions are 1.8 and 8.2.

����2x2 + x ± 15 = 0

62/87,21���For this equation, a = 2, b = 1, and c = ±15.

� The solutions are 2.5 and ±3.

����%$6(%$//��Elias hits a baseball into the air. The equation h = ±16t2 + 60t +3 models the height h in feet of the ball after t seconds. How long is the ball in the air?

62/87,21���

� The ball is in the air for about 3.8 seconds.

����Graph {(í2, 4), (í1, 1), (0, 0), (1, 1), (2, 4)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

62/87,21���

The points are reflected across a line of symmetry. They represent a quadratic function.

����Look for a pattern in the table to determine which kind of model best describes the data.

62/87,21���

� Calculate the first differences.

6LQFH�WKH�ILUVW�GLIIHUHQFHV�DUH�HTXDO��D�OLQHDU�IXQFWLRQ�PRGHOV�WKH�GDWD��Find the slope and y -LQWHUFHSW��

� Use the slope and one point from the table to find the y -LQWHUFHSW��

The data describes the linear function f (x) = 2x + 1.

����CAR CLUB The table shows the number of car club members for four consecutive years after it began. �

� a. Determine which model best represents the data. b. Write a function that models the data. c. Predict the number of car club members after 6 years.

62/87,21���a. First compare the first differences in the number of members:

� Since they're not equal, compare the second differences:

� Since the second differences are not equal, look for a common ratio:

� 7KHUH�LV�D�FRPPRQ�UDWLR�RI����VR�WKH�EHVW�PRGHO�ZRXOG�EH�DQ�H[SRQHQWLDO�IXQFWLRQ�WR�ILW�WKH�GDWD�� � b. A function that fits the data should be an exponential function with a ratio of 2. We are looking for a function of the form . Use a point from the data given to determine the value of a.

� � The function that best fits the data is:

� c. To find the number of car club members after 6 years, use the function found in part b and plug in 6 for x.

7KHUH�ZLOO�EH�����PHPEHUV�DIWHU���\HDUV���

Graph each function.����f (x) = |x í 1|

62/87,21���Make a table of values. �

�

x f (x) ±3 4 ±2 3 ±1 2 0 1 1 0 2 1 3 2 4 3

����f (x) = í|2x|

62/87,21���The negative in front of the absolute value makes the graph point down. Make a table of values.�

�

x f (x) ±2 ±4 ±1 ±2 0 0 1 ±2 2 ±4

����f (x) =

62/87,21���Make a table of values. �

�

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

����f (x) =

62/87,21���This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. �

�

x f (x) ±2 ±5 ±1 ±3 0 ±1 1 1 2 ±1 3 0 4 1

����Determine the domain and range of the function graphed below.

62/87,21���The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = �í 2, so the domain is all real numbers and the range is {y | y �í 2}.

eSolutions Manual - Powered by Cognero Page 5

Practice Test - Chapter 9

Use a table of values to graph the following functions. State the domain and range.���y = x2 + 2x + 5

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The graph opens upwards, so the vertex is a minimum located at (í1, 4). So the range is all real numbers greater than or equal to the minimum value of 4, or R = {y | y ����`�

x y �í3 8 í2 5 í1 4 0 5 1 8 2 13

���y = 2x2 í 3x + 1

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The range is all real numbers greater than or equal to the minimum value.

R = {y | y �� í0.125}

x y í2 15 í1 6 0 1 1 0 2 3 3 10

Consider y = x2 í 7x + 6.���Determine whether the function has a maximum or minimum value.

62/87,21���For y = x2 ± 7x + 6, a = 1, b = ±7, and c = 6. Because a is positive, the graph opens upward, so the function has a minimum value.

���State the maximum or minimum value.

62/87,21���Method 1: Find the vertex of the parabola. First, find the x-coordinate of the vertex. For the function y = x2 ± 7x + 6, use a = 1, b = ±7, and c = 6. �

� Substitute 3.5 for x in the function to find the value of the y -FRRUGLQDWH� �

� So, the minimum value is ±6.25. Method 2: Use a graphing calculator to graph the related function f (x) = x2 ± 7x + 6. Use the minimum option on the 2nd [CALC] menu to determine the minimum value.

Therefore, the minimum value is ±6.25.

���What are the domain and range?

62/87,21���Any number can be substituted for x in the function. So, tKH�GRPDLQ�LV�'� �^DOO�UHDO�QXPEHUV`�� The minimum of the function is í6.25. Therefore, R = {y | y �� í6.25} because the range is all real numbers that are JUHDWHU�WKDQ�RU�HTXDO�WR�WKH�PLQLPXP�YDOXH��

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

���x2 + 7x + 10 = 0

62/87,21���Graph the related function f (x) = x2 + 7x + 10.

The x-intercepts of the graph appear to be at ±5 and ±2, so the solutions are ±5 and ±2.Check:

and

���x2 í 5 = í3x

62/87,21���Write the equation in standard form.

Graph the related function f (x) = x2 + 3x í 5.

The x-intercepts are located between ±5 and ±4 and between 1 and 2. Make a table using an increment of 0.1 for the x-values located between ±5 and ±4 and between 1 and 2.

�

For each table, the function value that is closest to zero when the sign changes is 0.04. Thus, the roots are approximately ±4.2 and 1.2.

x ±4.5 ±4.4 ±4.3 ±4.2 ±4.1 y 1.75 1.16 0.59 0.04 ±0.49

x 1.1 1.2 1.3 1.4 1.5 y ±0.49 0.04 0.59 1.16 1.75

Describe how the graph of each function is related to the graph of f (x) = x2.���g(x) = x2 í 5

62/87,21���

The graph of f (x) = x2 + c represents a vertical translation of the parent graph. The value of c is ±5, and ±5 < 0. If c < 0, the graph of f (x) = x2 is translated �XQLWV�GRZQ��7KHUHIRUH��WKH�JUDSK�RI�y = x2 ± 5 is a translation of the graph

of y = x2 shifted down 5 units.

���g(x) = í3x2

62/87,21���

�7KH�JUDSK�RI�f (x) = ax2 expands or compresses the parent graph vertically. Since |a| > 1, the graph is vertically expanded. Since a < 1, the graph is reflected across the x-axis. So, the graph of y = í3x2 is the graph of y = x2 reflected across the x-axis and vertically expanded. �

����

62/87,21���

The function can be written f (x) = ax2 + c, where a = �DQG�c = 4. Since 4 > 0 and 0 < ������WKH�JUDSK�RI�y =

x2 + 4 is the graph of y = x2 vertically compressed and shifted up 4 units.

����08/7,3/(�&+2,&(� Which is an equation for the function shown in the graph?

$� y = í3x2 %� y = 3x2 + 1 &� y = x2 + 2 '� y = í3x2 + 2

62/87,21���The parabola opens downward, so the a in the equation must be negative. Therefore, choices B and C can be eliminated. The y-intercept is at the point (0, c), or in this case (0, 2). Therefore c = 2, so choice A can be eliminated.So, the correct choice is choice D.

Solve each equation by completing the square.����x2 + 2x + 5 = 0

62/87,21���

� No real number has a negative square. So, this equation has no real solutions.

����x2 í x í 6 = 0

62/87,21���

� The solutions are 3 and ±2.

����2x2 í 36 = í6x

62/87,21���Rewrite the equation in standard form.

� The solutions are 3 and ±6.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.����x2 í x í 30 = 0

62/87,21���For this equation, a = 1, b = ±1, and c = ±30.

� The solutions are 6 and ±5.

����x2 í 10x = í15

62/87,21���Rewrite the equation in standard form. �

� For this equation, a = 1, b = ±10, and c = 15.

� The solutions are 1.8 and 8.2.

����2x2 + x ± 15 = 0

62/87,21���For this equation, a = 2, b = 1, and c = ±15.

� The solutions are 2.5 and ±3.

����%$6(%$//��Elias hits a baseball into the air. The equation h = ±16t2 + 60t +3 models the height h in feet of the ball after t seconds. How long is the ball in the air?

62/87,21���

� The ball is in the air for about 3.8 seconds.

����Graph {(í2, 4), (í1, 1), (0, 0), (1, 1), (2, 4)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

62/87,21���

The points are reflected across a line of symmetry. They represent a quadratic function.

����Look for a pattern in the table to determine which kind of model best describes the data.

62/87,21���

� Calculate the first differences.

6LQFH�WKH�ILUVW�GLIIHUHQFHV�DUH�HTXDO��D�OLQHDU�IXQFWLRQ�PRGHOV�WKH�GDWD��Find the slope and y -LQWHUFHSW��

� Use the slope and one point from the table to find the y -LQWHUFHSW��

The data describes the linear function f (x) = 2x + 1.

����CAR CLUB The table shows the number of car club members for four consecutive years after it began. �

� a. Determine which model best represents the data. b. Write a function that models the data. c. Predict the number of car club members after 6 years.

62/87,21���a. First compare the first differences in the number of members:

� Since they're not equal, compare the second differences:

� Since the second differences are not equal, look for a common ratio:

� 7KHUH�LV�D�FRPPRQ�UDWLR�RI����VR�WKH�EHVW�PRGHO�ZRXOG�EH�DQ�H[SRQHQWLDO�IXQFWLRQ�WR�ILW�WKH�GDWD�� � b. A function that fits the data should be an exponential function with a ratio of 2. We are looking for a function of the form . Use a point from the data given to determine the value of a.

� � The function that best fits the data is:

� c. To find the number of car club members after 6 years, use the function found in part b and plug in 6 for x.

7KHUH�ZLOO�EH�����PHPEHUV�DIWHU���\HDUV���

Graph each function.����f (x) = |x í 1|

62/87,21���Make a table of values. �

�

x f (x) ±3 4 ±2 3 ±1 2 0 1 1 0 2 1 3 2 4 3

����f (x) = í|2x|

62/87,21���The negative in front of the absolute value makes the graph point down. Make a table of values.�

�

x f (x) ±2 ±4 ±1 ±2 0 0 1 ±2 2 ±4

����f (x) =

62/87,21���Make a table of values. �

�

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

����f (x) =

62/87,21���This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. �

�

x f (x) ±2 ±5 ±1 ±3 0 ±1 1 1 2 ±1 3 0 4 1

����Determine the domain and range of the function graphed below.

62/87,21���The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = �í 2, so the domain is all real numbers and the range is {y | y �í 2}.

eSolutions Manual - Powered by Cognero Page 6

Practice Test - Chapter 9

Use a table of values to graph the following functions. State the domain and range.���y = x2 + 2x + 5

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The graph opens upwards, so the vertex is a minimum located at (í1, 4). So the range is all real numbers greater than or equal to the minimum value of 4, or R = {y | y ����`�

x y �í3 8 í2 5 í1 4 0 5 1 8 2 13

���y = 2x2 í 3x + 1

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The range is all real numbers greater than or equal to the minimum value.

R = {y | y �� í0.125}

x y í2 15 í1 6 0 1 1 0 2 3 3 10

Consider y = x2 í 7x + 6.���Determine whether the function has a maximum or minimum value.

62/87,21���For y = x2 ± 7x + 6, a = 1, b = ±7, and c = 6. Because a is positive, the graph opens upward, so the function has a minimum value.

���State the maximum or minimum value.

62/87,21���Method 1: Find the vertex of the parabola. First, find the x-coordinate of the vertex. For the function y = x2 ± 7x + 6, use a = 1, b = ±7, and c = 6. �

� Substitute 3.5 for x in the function to find the value of the y -FRRUGLQDWH� �

� So, the minimum value is ±6.25. Method 2: Use a graphing calculator to graph the related function f (x) = x2 ± 7x + 6. Use the minimum option on the 2nd [CALC] menu to determine the minimum value.

Therefore, the minimum value is ±6.25.

���What are the domain and range?

62/87,21���Any number can be substituted for x in the function. So, tKH�GRPDLQ�LV�'� �^DOO�UHDO�QXPEHUV`�� The minimum of the function is í6.25. Therefore, R = {y | y �� í6.25} because the range is all real numbers that are JUHDWHU�WKDQ�RU�HTXDO�WR�WKH�PLQLPXP�YDOXH��

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

���x2 + 7x + 10 = 0

62/87,21���Graph the related function f (x) = x2 + 7x + 10.

The x-intercepts of the graph appear to be at ±5 and ±2, so the solutions are ±5 and ±2.Check:

and

���x2 í 5 = í3x

62/87,21���Write the equation in standard form.

Graph the related function f (x) = x2 + 3x í 5.

The x-intercepts are located between ±5 and ±4 and between 1 and 2. Make a table using an increment of 0.1 for the x-values located between ±5 and ±4 and between 1 and 2.

�

For each table, the function value that is closest to zero when the sign changes is 0.04. Thus, the roots are approximately ±4.2 and 1.2.

x ±4.5 ±4.4 ±4.3 ±4.2 ±4.1 y 1.75 1.16 0.59 0.04 ±0.49

x 1.1 1.2 1.3 1.4 1.5 y ±0.49 0.04 0.59 1.16 1.75

Describe how the graph of each function is related to the graph of f (x) = x2.���g(x) = x2 í 5

62/87,21���

The graph of f (x) = x2 + c represents a vertical translation of the parent graph. The value of c is ±5, and ±5 < 0. If c < 0, the graph of f (x) = x2 is translated �XQLWV�GRZQ��7KHUHIRUH��WKH�JUDSK�RI�y = x2 ± 5 is a translation of the graph

of y = x2 shifted down 5 units.

���g(x) = í3x2

62/87,21���

�7KH�JUDSK�RI�f (x) = ax2 expands or compresses the parent graph vertically. Since |a| > 1, the graph is vertically expanded. Since a < 1, the graph is reflected across the x-axis. So, the graph of y = í3x2 is the graph of y = x2 reflected across the x-axis and vertically expanded. �

����

62/87,21���

The function can be written f (x) = ax2 + c, where a = �DQG�c = 4. Since 4 > 0 and 0 < ������WKH�JUDSK�RI�y =

x2 + 4 is the graph of y = x2 vertically compressed and shifted up 4 units.

����08/7,3/(�&+2,&(� Which is an equation for the function shown in the graph?

$� y = í3x2 %� y = 3x2 + 1 &� y = x2 + 2 '� y = í3x2 + 2

62/87,21���The parabola opens downward, so the a in the equation must be negative. Therefore, choices B and C can be eliminated. The y-intercept is at the point (0, c), or in this case (0, 2). Therefore c = 2, so choice A can be eliminated.So, the correct choice is choice D.

Solve each equation by completing the square.����x2 + 2x + 5 = 0

62/87,21���

� No real number has a negative square. So, this equation has no real solutions.

����x2 í x í 6 = 0

62/87,21���

� The solutions are 3 and ±2.

����2x2 í 36 = í6x

62/87,21���Rewrite the equation in standard form.

� The solutions are 3 and ±6.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.����x2 í x í 30 = 0

62/87,21���For this equation, a = 1, b = ±1, and c = ±30.

� The solutions are 6 and ±5.

����x2 í 10x = í15

62/87,21���Rewrite the equation in standard form. �

� For this equation, a = 1, b = ±10, and c = 15.

� The solutions are 1.8 and 8.2.

����2x2 + x ± 15 = 0

62/87,21���For this equation, a = 2, b = 1, and c = ±15.

� The solutions are 2.5 and ±3.

����%$6(%$//��Elias hits a baseball into the air. The equation h = ±16t2 + 60t +3 models the height h in feet of the ball after t seconds. How long is the ball in the air?

62/87,21���

� The ball is in the air for about 3.8 seconds.

����Graph {(í2, 4), (í1, 1), (0, 0), (1, 1), (2, 4)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

62/87,21���

The points are reflected across a line of symmetry. They represent a quadratic function.

����Look for a pattern in the table to determine which kind of model best describes the data.

62/87,21���

� Calculate the first differences.

6LQFH�WKH�ILUVW�GLIIHUHQFHV�DUH�HTXDO��D�OLQHDU�IXQFWLRQ�PRGHOV�WKH�GDWD��Find the slope and y -LQWHUFHSW��

� Use the slope and one point from the table to find the y -LQWHUFHSW��

The data describes the linear function f (x) = 2x + 1.

����CAR CLUB The table shows the number of car club members for four consecutive years after it began. �

� a. Determine which model best represents the data. b. Write a function that models the data. c. Predict the number of car club members after 6 years.

62/87,21���a. First compare the first differences in the number of members:

� Since they're not equal, compare the second differences:

� Since the second differences are not equal, look for a common ratio:

� 7KHUH�LV�D�FRPPRQ�UDWLR�RI����VR�WKH�EHVW�PRGHO�ZRXOG�EH�DQ�H[SRQHQWLDO�IXQFWLRQ�WR�ILW�WKH�GDWD�� � b. A function that fits the data should be an exponential function with a ratio of 2. We are looking for a function of the form . Use a point from the data given to determine the value of a.

� � The function that best fits the data is:

� c. To find the number of car club members after 6 years, use the function found in part b and plug in 6 for x.

7KHUH�ZLOO�EH�����PHPEHUV�DIWHU���\HDUV���

Graph each function.����f (x) = |x í 1|

62/87,21���Make a table of values. �

�

x f (x) ±3 4 ±2 3 ±1 2 0 1 1 0 2 1 3 2 4 3

����f (x) = í|2x|

62/87,21���The negative in front of the absolute value makes the graph point down. Make a table of values.�

�

x f (x) ±2 ±4 ±1 ±2 0 0 1 ±2 2 ±4

����f (x) =

62/87,21���Make a table of values. �

�

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

����f (x) =

62/87,21���This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. �

�

x f (x) ±2 ±5 ±1 ±3 0 ±1 1 1 2 ±1 3 0 4 1

����Determine the domain and range of the function graphed below.

62/87,21���The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = �í 2, so the domain is all real numbers and the range is {y | y �í 2}.

eSolutions Manual - Powered by Cognero Page 7

Practice Test - Chapter 9

Use a table of values to graph the following functions. State the domain and range.���y = x2 + 2x + 5

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The graph opens upwards, so the vertex is a minimum located at (í1, 4). So the range is all real numbers greater than or equal to the minimum value of 4, or R = {y | y ����`�

x y �í3 8 í2 5 í1 4 0 5 1 8 2 13

���y = 2x2 í 3x + 1

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The range is all real numbers greater than or equal to the minimum value.

R = {y | y �� í0.125}

x y í2 15 í1 6 0 1 1 0 2 3 3 10

Consider y = x2 í 7x + 6.���Determine whether the function has a maximum or minimum value.

62/87,21���For y = x2 ± 7x + 6, a = 1, b = ±7, and c = 6. Because a is positive, the graph opens upward, so the function has a minimum value.

���State the maximum or minimum value.

62/87,21���Method 1: Find the vertex of the parabola. First, find the x-coordinate of the vertex. For the function y = x2 ± 7x + 6, use a = 1, b = ±7, and c = 6. �

� Substitute 3.5 for x in the function to find the value of the y -FRRUGLQDWH� �

� So, the minimum value is ±6.25. Method 2: Use a graphing calculator to graph the related function f (x) = x2 ± 7x + 6. Use the minimum option on the 2nd [CALC] menu to determine the minimum value.

Therefore, the minimum value is ±6.25.

���What are the domain and range?

62/87,21���Any number can be substituted for x in the function. So, tKH�GRPDLQ�LV�'� �^DOO�UHDO�QXPEHUV`�� The minimum of the function is í6.25. Therefore, R = {y | y �� í6.25} because the range is all real numbers that are JUHDWHU�WKDQ�RU�HTXDO�WR�WKH�PLQLPXP�YDOXH��

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

���x2 + 7x + 10 = 0

62/87,21���Graph the related function f (x) = x2 + 7x + 10.

The x-intercepts of the graph appear to be at ±5 and ±2, so the solutions are ±5 and ±2.Check:

and

���x2 í 5 = í3x

62/87,21���Write the equation in standard form.

Graph the related function f (x) = x2 + 3x í 5.

The x-intercepts are located between ±5 and ±4 and between 1 and 2. Make a table using an increment of 0.1 for the x-values located between ±5 and ±4 and between 1 and 2.

�

For each table, the function value that is closest to zero when the sign changes is 0.04. Thus, the roots are approximately ±4.2 and 1.2.

x ±4.5 ±4.4 ±4.3 ±4.2 ±4.1 y 1.75 1.16 0.59 0.04 ±0.49

x 1.1 1.2 1.3 1.4 1.5 y ±0.49 0.04 0.59 1.16 1.75

Describe how the graph of each function is related to the graph of f (x) = x2.���g(x) = x2 í 5

62/87,21���

The graph of f (x) = x2 + c represents a vertical translation of the parent graph. The value of c is ±5, and ±5 < 0. If c < 0, the graph of f (x) = x2 is translated �XQLWV�GRZQ��7KHUHIRUH��WKH�JUDSK�RI�y = x2 ± 5 is a translation of the graph

of y = x2 shifted down 5 units.

���g(x) = í3x2

62/87,21���

�7KH�JUDSK�RI�f (x) = ax2 expands or compresses the parent graph vertically. Since |a| > 1, the graph is vertically expanded. Since a < 1, the graph is reflected across the x-axis. So, the graph of y = í3x2 is the graph of y = x2 reflected across the x-axis and vertically expanded. �

����

62/87,21���

The function can be written f (x) = ax2 + c, where a = �DQG�c = 4. Since 4 > 0 and 0 < ������WKH�JUDSK�RI�y =

x2 + 4 is the graph of y = x2 vertically compressed and shifted up 4 units.

����08/7,3/(�&+2,&(� Which is an equation for the function shown in the graph?

$� y = í3x2 %� y = 3x2 + 1 &� y = x2 + 2 '� y = í3x2 + 2

62/87,21���The parabola opens downward, so the a in the equation must be negative. Therefore, choices B and C can be eliminated. The y-intercept is at the point (0, c), or in this case (0, 2). Therefore c = 2, so choice A can be eliminated.So, the correct choice is choice D.

Solve each equation by completing the square.����x2 + 2x + 5 = 0

62/87,21���

� No real number has a negative square. So, this equation has no real solutions.

����x2 í x í 6 = 0

62/87,21���

� The solutions are 3 and ±2.

����2x2 í 36 = í6x

62/87,21���Rewrite the equation in standard form.

� The solutions are 3 and ±6.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.����x2 í x í 30 = 0

62/87,21���For this equation, a = 1, b = ±1, and c = ±30.

� The solutions are 6 and ±5.

����x2 í 10x = í15

62/87,21���Rewrite the equation in standard form. �

� For this equation, a = 1, b = ±10, and c = 15.

� The solutions are 1.8 and 8.2.

����2x2 + x ± 15 = 0

62/87,21���For this equation, a = 2, b = 1, and c = ±15.

� The solutions are 2.5 and ±3.

����%$6(%$//��Elias hits a baseball into the air. The equation h = ±16t2 + 60t +3 models the height h in feet of the ball after t seconds. How long is the ball in the air?

62/87,21���

� The ball is in the air for about 3.8 seconds.

����Graph {(í2, 4), (í1, 1), (0, 0), (1, 1), (2, 4)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

62/87,21���

The points are reflected across a line of symmetry. They represent a quadratic function.

����Look for a pattern in the table to determine which kind of model best describes the data.

62/87,21���

� Calculate the first differences.

6LQFH�WKH�ILUVW�GLIIHUHQFHV�DUH�HTXDO��D�OLQHDU�IXQFWLRQ�PRGHOV�WKH�GDWD��Find the slope and y -LQWHUFHSW��

� Use the slope and one point from the table to find the y -LQWHUFHSW��

The data describes the linear function f (x) = 2x + 1.

����CAR CLUB The table shows the number of car club members for four consecutive years after it began. �

� a. Determine which model best represents the data. b. Write a function that models the data. c. Predict the number of car club members after 6 years.

62/87,21���a. First compare the first differences in the number of members:

� Since they're not equal, compare the second differences:

� Since the second differences are not equal, look for a common ratio:

� 7KHUH�LV�D�FRPPRQ�UDWLR�RI����VR�WKH�EHVW�PRGHO�ZRXOG�EH�DQ�H[SRQHQWLDO�IXQFWLRQ�WR�ILW�WKH�GDWD�� � b. A function that fits the data should be an exponential function with a ratio of 2. We are looking for a function of the form . Use a point from the data given to determine the value of a.

� � The function that best fits the data is:

� c. To find the number of car club members after 6 years, use the function found in part b and plug in 6 for x.

7KHUH�ZLOO�EH�����PHPEHUV�DIWHU���\HDUV���

Graph each function.����f (x) = |x í 1|

62/87,21���Make a table of values. �

�

x f (x) ±3 4 ±2 3 ±1 2 0 1 1 0 2 1 3 2 4 3

����f (x) = í|2x|

62/87,21���The negative in front of the absolute value makes the graph point down. Make a table of values.�

�

x f (x) ±2 ±4 ±1 ±2 0 0 1 ±2 2 ±4

����f (x) =

62/87,21���Make a table of values. �

�

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

����f (x) =

62/87,21���This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. �

�

x f (x) ±2 ±5 ±1 ±3 0 ±1 1 1 2 ±1 3 0 4 1

����Determine the domain and range of the function graphed below.

62/87,21���The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = �í 2, so the domain is all real numbers and the range is {y | y �í 2}.

eSolutions Manual - Powered by Cognero Page 8

Practice Test - Chapter 9

Use a table of values to graph the following functions. State the domain and range.���y = x2 + 2x + 5

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The graph opens upwards, so the vertex is a minimum located at (í1, 4). So the range is all real numbers greater than or equal to the minimum value of 4, or R = {y | y ����`�

x y �í3 8 í2 5 í1 4 0 5 1 8 2 13

���y = 2x2 í 3x + 1

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The range is all real numbers greater than or equal to the minimum value.

R = {y | y �� í0.125}

x y í2 15 í1 6 0 1 1 0 2 3 3 10

Consider y = x2 í 7x + 6.���Determine whether the function has a maximum or minimum value.

62/87,21���For y = x2 ± 7x + 6, a = 1, b = ±7, and c = 6. Because a is positive, the graph opens upward, so the function has a minimum value.

���State the maximum or minimum value.

62/87,21���Method 1: Find the vertex of the parabola. First, find the x-coordinate of the vertex. For the function y = x2 ± 7x + 6, use a = 1, b = ±7, and c = 6. �

� Substitute 3.5 for x in the function to find the value of the y -FRRUGLQDWH� �

� So, the minimum value is ±6.25. Method 2: Use a graphing calculator to graph the related function f (x) = x2 ± 7x + 6. Use the minimum option on the 2nd [CALC] menu to determine the minimum value.

Therefore, the minimum value is ±6.25.

���What are the domain and range?

62/87,21���Any number can be substituted for x in the function. So, tKH�GRPDLQ�LV�'� �^DOO�UHDO�QXPEHUV`�� The minimum of the function is í6.25. Therefore, R = {y | y �� í6.25} because the range is all real numbers that are JUHDWHU�WKDQ�RU�HTXDO�WR�WKH�PLQLPXP�YDOXH��

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

���x2 + 7x + 10 = 0

62/87,21���Graph the related function f (x) = x2 + 7x + 10.

The x-intercepts of the graph appear to be at ±5 and ±2, so the solutions are ±5 and ±2.Check:

and

���x2 í 5 = í3x

62/87,21���Write the equation in standard form.

Graph the related function f (x) = x2 + 3x í 5.

The x-intercepts are located between ±5 and ±4 and between 1 and 2. Make a table using an increment of 0.1 for the x-values located between ±5 and ±4 and between 1 and 2.

�

For each table, the function value that is closest to zero when the sign changes is 0.04. Thus, the roots are approximately ±4.2 and 1.2.

x ±4.5 ±4.4 ±4.3 ±4.2 ±4.1 y 1.75 1.16 0.59 0.04 ±0.49

x 1.1 1.2 1.3 1.4 1.5 y ±0.49 0.04 0.59 1.16 1.75

Describe how the graph of each function is related to the graph of f (x) = x2.���g(x) = x2 í 5

62/87,21���

The graph of f (x) = x2 + c represents a vertical translation of the parent graph. The value of c is ±5, and ±5 < 0. If c < 0, the graph of f (x) = x2 is translated �XQLWV�GRZQ��7KHUHIRUH��WKH�JUDSK�RI�y = x2 ± 5 is a translation of the graph

of y = x2 shifted down 5 units.

���g(x) = í3x2

62/87,21���

�7KH�JUDSK�RI�f (x) = ax2 expands or compresses the parent graph vertically. Since |a| > 1, the graph is vertically expanded. Since a < 1, the graph is reflected across the x-axis. So, the graph of y = í3x2 is the graph of y = x2 reflected across the x-axis and vertically expanded. �

����

62/87,21���

The function can be written f (x) = ax2 + c, where a = �DQG�c = 4. Since 4 > 0 and 0 < ������WKH�JUDSK�RI�y =

x2 + 4 is the graph of y = x2 vertically compressed and shifted up 4 units.

����08/7,3/(�&+2,&(� Which is an equation for the function shown in the graph?

$� y = í3x2 %� y = 3x2 + 1 &� y = x2 + 2 '� y = í3x2 + 2

62/87,21���The parabola opens downward, so the a in the equation must be negative. Therefore, choices B and C can be eliminated. The y-intercept is at the point (0, c), or in this case (0, 2). Therefore c = 2, so choice A can be eliminated.So, the correct choice is choice D.

Solve each equation by completing the square.����x2 + 2x + 5 = 0

62/87,21���

� No real number has a negative square. So, this equation has no real solutions.

����x2 í x í 6 = 0

62/87,21���

� The solutions are 3 and ±2.

����2x2 í 36 = í6x

62/87,21���Rewrite the equation in standard form.

� The solutions are 3 and ±6.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.����x2 í x í 30 = 0

62/87,21���For this equation, a = 1, b = ±1, and c = ±30.

� The solutions are 6 and ±5.

����x2 í 10x = í15

62/87,21���Rewrite the equation in standard form. �

� For this equation, a = 1, b = ±10, and c = 15.

� The solutions are 1.8 and 8.2.

����2x2 + x ± 15 = 0

62/87,21���For this equation, a = 2, b = 1, and c = ±15.

� The solutions are 2.5 and ±3.

����%$6(%$//��Elias hits a baseball into the air. The equation h = ±16t2 + 60t +3 models the height h in feet of the ball after t seconds. How long is the ball in the air?

62/87,21���

� The ball is in the air for about 3.8 seconds.

����Graph {(í2, 4), (í1, 1), (0, 0), (1, 1), (2, 4)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

62/87,21���

The points are reflected across a line of symmetry. They represent a quadratic function.

����Look for a pattern in the table to determine which kind of model best describes the data.

62/87,21���

� Calculate the first differences.

6LQFH�WKH�ILUVW�GLIIHUHQFHV�DUH�HTXDO��D�OLQHDU�IXQFWLRQ�PRGHOV�WKH�GDWD��Find the slope and y -LQWHUFHSW��

� Use the slope and one point from the table to find the y -LQWHUFHSW��

The data describes the linear function f (x) = 2x + 1.

����CAR CLUB The table shows the number of car club members for four consecutive years after it began. �

� a. Determine which model best represents the data. b. Write a function that models the data. c. Predict the number of car club members after 6 years.

62/87,21���a. First compare the first differences in the number of members:

� Since they're not equal, compare the second differences:

� Since the second differences are not equal, look for a common ratio:

� 7KHUH�LV�D�FRPPRQ�UDWLR�RI����VR�WKH�EHVW�PRGHO�ZRXOG�EH�DQ�H[SRQHQWLDO�IXQFWLRQ�WR�ILW�WKH�GDWD�� � b. A function that fits the data should be an exponential function with a ratio of 2. We are looking for a function of the form . Use a point from the data given to determine the value of a.

� � The function that best fits the data is:

� c. To find the number of car club members after 6 years, use the function found in part b and plug in 6 for x.

7KHUH�ZLOO�EH�����PHPEHUV�DIWHU���\HDUV���

Graph each function.����f (x) = |x í 1|

62/87,21���Make a table of values. �

�

x f (x) ±3 4 ±2 3 ±1 2 0 1 1 0 2 1 3 2 4 3

����f (x) = í|2x|

62/87,21���The negative in front of the absolute value makes the graph point down. Make a table of values.�

�

x f (x) ±2 ±4 ±1 ±2 0 0 1 ±2 2 ±4

����f (x) =

62/87,21���Make a table of values. �

�

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

����f (x) =

62/87,21���This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. �

�

x f (x) ±2 ±5 ±1 ±3 0 ±1 1 1 2 ±1 3 0 4 1

����Determine the domain and range of the function graphed below.

62/87,21���The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = �í 2, so the domain is all real numbers and the range is {y | y �í 2}.

eSolutions Manual - Powered by Cognero Page 9

Practice Test - Chapter 9

Use a table of values to graph the following functions. State the domain and range.���y = x2 + 2x + 5

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The graph opens upwards, so the vertex is a minimum located at (í1, 4). So the range is all real numbers greater than or equal to the minimum value of 4, or R = {y | y ����`�

x y �í3 8 í2 5 í1 4 0 5 1 8 2 13

���y = 2x2 í 3x + 1

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The range is all real numbers greater than or equal to the minimum value.

R = {y | y �� í0.125}

x y í2 15 í1 6 0 1 1 0 2 3 3 10

Consider y = x2 í 7x + 6.���Determine whether the function has a maximum or minimum value.

62/87,21���For y = x2 ± 7x + 6, a = 1, b = ±7, and c = 6. Because a is positive, the graph opens upward, so the function has a minimum value.

���State the maximum or minimum value.

62/87,21���Method 1: Find the vertex of the parabola. First, find the x-coordinate of the vertex. For the function y = x2 ± 7x + 6, use a = 1, b = ±7, and c = 6. �

� Substitute 3.5 for x in the function to find the value of the y -FRRUGLQDWH� �

� So, the minimum value is ±6.25. Method 2: Use a graphing calculator to graph the related function f (x) = x2 ± 7x + 6. Use the minimum option on the 2nd [CALC] menu to determine the minimum value.

Therefore, the minimum value is ±6.25.

���What are the domain and range?

62/87,21���Any number can be substituted for x in the function. So, tKH�GRPDLQ�LV�'� �^DOO�UHDO�QXPEHUV`�� The minimum of the function is í6.25. Therefore, R = {y | y �� í6.25} because the range is all real numbers that are JUHDWHU�WKDQ�RU�HTXDO�WR�WKH�PLQLPXP�YDOXH��

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

���x2 + 7x + 10 = 0

62/87,21���Graph the related function f (x) = x2 + 7x + 10.

The x-intercepts of the graph appear to be at ±5 and ±2, so the solutions are ±5 and ±2.Check:

and

���x2 í 5 = í3x

62/87,21���Write the equation in standard form.

Graph the related function f (x) = x2 + 3x í 5.

The x-intercepts are located between ±5 and ±4 and between 1 and 2. Make a table using an increment of 0.1 for the x-values located between ±5 and ±4 and between 1 and 2.

�

For each table, the function value that is closest to zero when the sign changes is 0.04. Thus, the roots are approximately ±4.2 and 1.2.

x ±4.5 ±4.4 ±4.3 ±4.2 ±4.1 y 1.75 1.16 0.59 0.04 ±0.49

x 1.1 1.2 1.3 1.4 1.5 y ±0.49 0.04 0.59 1.16 1.75

Describe how the graph of each function is related to the graph of f (x) = x2.���g(x) = x2 í 5

62/87,21���

The graph of f (x) = x2 + c represents a vertical translation of the parent graph. The value of c is ±5, and ±5 < 0. If c < 0, the graph of f (x) = x2 is translated �XQLWV�GRZQ��7KHUHIRUH��WKH�JUDSK�RI�y = x2 ± 5 is a translation of the graph

of y = x2 shifted down 5 units.

���g(x) = í3x2

62/87,21���

�7KH�JUDSK�RI�f (x) = ax2 expands or compresses the parent graph vertically. Since |a| > 1, the graph is vertically expanded. Since a < 1, the graph is reflected across the x-axis. So, the graph of y = í3x2 is the graph of y = x2 reflected across the x-axis and vertically expanded. �

����

62/87,21���

The function can be written f (x) = ax2 + c, where a = �DQG�c = 4. Since 4 > 0 and 0 < ������WKH�JUDSK�RI�y =

x2 + 4 is the graph of y = x2 vertically compressed and shifted up 4 units.

����08/7,3/(�&+2,&(� Which is an equation for the function shown in the graph?

$� y = í3x2 %� y = 3x2 + 1 &� y = x2 + 2 '� y = í3x2 + 2

62/87,21���The parabola opens downward, so the a in the equation must be negative. Therefore, choices B and C can be eliminated. The y-intercept is at the point (0, c), or in this case (0, 2). Therefore c = 2, so choice A can be eliminated.So, the correct choice is choice D.

Solve each equation by completing the square.����x2 + 2x + 5 = 0

62/87,21���

� No real number has a negative square. So, this equation has no real solutions.

����x2 í x í 6 = 0

62/87,21���

� The solutions are 3 and ±2.

����2x2 í 36 = í6x

62/87,21���Rewrite the equation in standard form.

� The solutions are 3 and ±6.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.����x2 í x í 30 = 0

62/87,21���For this equation, a = 1, b = ±1, and c = ±30.

� The solutions are 6 and ±5.

����x2 í 10x = í15

62/87,21���Rewrite the equation in standard form. �

� For this equation, a = 1, b = ±10, and c = 15.

� The solutions are 1.8 and 8.2.

����2x2 + x ± 15 = 0

62/87,21���For this equation, a = 2, b = 1, and c = ±15.

� The solutions are 2.5 and ±3.

����%$6(%$//��Elias hits a baseball into the air. The equation h = ±16t2 + 60t +3 models the height h in feet of the ball after t seconds. How long is the ball in the air?

62/87,21���

� The ball is in the air for about 3.8 seconds.

����Graph {(í2, 4), (í1, 1), (0, 0), (1, 1), (2, 4)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

62/87,21���

The points are reflected across a line of symmetry. They represent a quadratic function.

����Look for a pattern in the table to determine which kind of model best describes the data.

62/87,21���

� Calculate the first differences.

6LQFH�WKH�ILUVW�GLIIHUHQFHV�DUH�HTXDO��D�OLQHDU�IXQFWLRQ�PRGHOV�WKH�GDWD��Find the slope and y -LQWHUFHSW��

� Use the slope and one point from the table to find the y -LQWHUFHSW��

The data describes the linear function f (x) = 2x + 1.

����CAR CLUB The table shows the number of car club members for four consecutive years after it began. �

� a. Determine which model best represents the data. b. Write a function that models the data. c. Predict the number of car club members after 6 years.

62/87,21���a. First compare the first differences in the number of members:

� Since they're not equal, compare the second differences:

� Since the second differences are not equal, look for a common ratio:

� 7KHUH�LV�D�FRPPRQ�UDWLR�RI����VR�WKH�EHVW�PRGHO�ZRXOG�EH�DQ�H[SRQHQWLDO�IXQFWLRQ�WR�ILW�WKH�GDWD�� � b. A function that fits the data should be an exponential function with a ratio of 2. We are looking for a function of the form . Use a point from the data given to determine the value of a.

� � The function that best fits the data is:

� c. To find the number of car club members after 6 years, use the function found in part b and plug in 6 for x.

7KHUH�ZLOO�EH�����PHPEHUV�DIWHU���\HDUV���

Graph each function.����f (x) = |x í 1|

62/87,21���Make a table of values. �

�

x f (x) ±3 4 ±2 3 ±1 2 0 1 1 0 2 1 3 2 4 3

����f (x) = í|2x|

62/87,21���The negative in front of the absolute value makes the graph point down. Make a table of values.�

�

x f (x) ±2 ±4 ±1 ±2 0 0 1 ±2 2 ±4

����f (x) =

62/87,21���Make a table of values. �

�

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

����f (x) =

62/87,21���This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. �

�

x f (x) ±2 ±5 ±1 ±3 0 ±1 1 1 2 ±1 3 0 4 1

����Determine the domain and range of the function graphed below.

62/87,21���The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = �í 2, so the domain is all real numbers and the range is {y | y �í 2}.

eSolutions Manual - Powered by Cognero Page 10

Practice Test - Chapter 9

Use a table of values to graph the following functions. State the domain and range.���y = x2 + 2x + 5

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The graph opens upwards, so the vertex is a minimum located at (í1, 4). So the range is all real numbers greater than or equal to the minimum value of 4, or R = {y | y ����`�

x y �í3 8 í2 5 í1 4 0 5 1 8 2 13

���y = 2x2 í 3x + 1

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The range is all real numbers greater than or equal to the minimum value.

R = {y | y �� í0.125}

x y í2 15 í1 6 0 1 1 0 2 3 3 10

Consider y = x2 í 7x + 6.���Determine whether the function has a maximum or minimum value.

62/87,21���For y = x2 ± 7x + 6, a = 1, b = ±7, and c = 6. Because a is positive, the graph opens upward, so the function has a minimum value.

���State the maximum or minimum value.

62/87,21���Method 1: Find the vertex of the parabola. First, find the x-coordinate of the vertex. For the function y = x2 ± 7x + 6, use a = 1, b = ±7, and c = 6. �

� Substitute 3.5 for x in the function to find the value of the y -FRRUGLQDWH� �

� So, the minimum value is ±6.25. Method 2: Use a graphing calculator to graph the related function f (x) = x2 ± 7x + 6. Use the minimum option on the 2nd [CALC] menu to determine the minimum value.

Therefore, the minimum value is ±6.25.

���What are the domain and range?

62/87,21���Any number can be substituted for x in the function. So, tKH�GRPDLQ�LV�'� �^DOO�UHDO�QXPEHUV`�� The minimum of the function is í6.25. Therefore, R = {y | y �� í6.25} because the range is all real numbers that are JUHDWHU�WKDQ�RU�HTXDO�WR�WKH�PLQLPXP�YDOXH��

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

���x2 + 7x + 10 = 0

62/87,21���Graph the related function f (x) = x2 + 7x + 10.

The x-intercepts of the graph appear to be at ±5 and ±2, so the solutions are ±5 and ±2.Check:

and

���x2 í 5 = í3x

62/87,21���Write the equation in standard form.

Graph the related function f (x) = x2 + 3x í 5.

The x-intercepts are located between ±5 and ±4 and between 1 and 2. Make a table using an increment of 0.1 for the x-values located between ±5 and ±4 and between 1 and 2.

�

For each table, the function value that is closest to zero when the sign changes is 0.04. Thus, the roots are approximately ±4.2 and 1.2.

x ±4.5 ±4.4 ±4.3 ±4.2 ±4.1 y 1.75 1.16 0.59 0.04 ±0.49

x 1.1 1.2 1.3 1.4 1.5 y ±0.49 0.04 0.59 1.16 1.75

Describe how the graph of each function is related to the graph of f (x) = x2.���g(x) = x2 í 5

62/87,21���

The graph of f (x) = x2 + c represents a vertical translation of the parent graph. The value of c is ±5, and ±5 < 0. If c < 0, the graph of f (x) = x2 is translated �XQLWV�GRZQ��7KHUHIRUH��WKH�JUDSK�RI�y = x2 ± 5 is a translation of the graph

of y = x2 shifted down 5 units.

���g(x) = í3x2

62/87,21���

�7KH�JUDSK�RI�f (x) = ax2 expands or compresses the parent graph vertically. Since |a| > 1, the graph is vertically expanded. Since a < 1, the graph is reflected across the x-axis. So, the graph of y = í3x2 is the graph of y = x2 reflected across the x-axis and vertically expanded. �

����

62/87,21���

The function can be written f (x) = ax2 + c, where a = �DQG�c = 4. Since 4 > 0 and 0 < ������WKH�JUDSK�RI�y =

x2 + 4 is the graph of y = x2 vertically compressed and shifted up 4 units.

����08/7,3/(�&+2,&(� Which is an equation for the function shown in the graph?

$� y = í3x2 %� y = 3x2 + 1 &� y = x2 + 2 '� y = í3x2 + 2

62/87,21���The parabola opens downward, so the a in the equation must be negative. Therefore, choices B and C can be eliminated. The y-intercept is at the point (0, c), or in this case (0, 2). Therefore c = 2, so choice A can be eliminated.So, the correct choice is choice D.

Solve each equation by completing the square.����x2 + 2x + 5 = 0

62/87,21���

� No real number has a negative square. So, this equation has no real solutions.

����x2 í x í 6 = 0

62/87,21���

� The solutions are 3 and ±2.

����2x2 í 36 = í6x

62/87,21���Rewrite the equation in standard form.

� The solutions are 3 and ±6.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.����x2 í x í 30 = 0

62/87,21���For this equation, a = 1, b = ±1, and c = ±30.

� The solutions are 6 and ±5.

����x2 í 10x = í15

62/87,21���Rewrite the equation in standard form. �

� For this equation, a = 1, b = ±10, and c = 15.

� The solutions are 1.8 and 8.2.

����2x2 + x ± 15 = 0

62/87,21���For this equation, a = 2, b = 1, and c = ±15.

� The solutions are 2.5 and ±3.

����%$6(%$//��Elias hits a baseball into the air. The equation h = ±16t2 + 60t +3 models the height h in feet of the ball after t seconds. How long is the ball in the air?

62/87,21���

� The ball is in the air for about 3.8 seconds.

����Graph {(í2, 4), (í1, 1), (0, 0), (1, 1), (2, 4)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

62/87,21���

The points are reflected across a line of symmetry. They represent a quadratic function.

����Look for a pattern in the table to determine which kind of model best describes the data.

62/87,21���

� Calculate the first differences.

6LQFH�WKH�ILUVW�GLIIHUHQFHV�DUH�HTXDO��D�OLQHDU�IXQFWLRQ�PRGHOV�WKH�GDWD��Find the slope and y -LQWHUFHSW��

� Use the slope and one point from the table to find the y -LQWHUFHSW��

The data describes the linear function f (x) = 2x + 1.

����CAR CLUB The table shows the number of car club members for four consecutive years after it began. �

� a. Determine which model best represents the data. b. Write a function that models the data. c. Predict the number of car club members after 6 years.

62/87,21���a. First compare the first differences in the number of members:

� Since they're not equal, compare the second differences:

� Since the second differences are not equal, look for a common ratio:

� 7KHUH�LV�D�FRPPRQ�UDWLR�RI����VR�WKH�EHVW�PRGHO�ZRXOG�EH�DQ�H[SRQHQWLDO�IXQFWLRQ�WR�ILW�WKH�GDWD�� � b. A function that fits the data should be an exponential function with a ratio of 2. We are looking for a function of the form . Use a point from the data given to determine the value of a.

� � The function that best fits the data is:

� c. To find the number of car club members after 6 years, use the function found in part b and plug in 6 for x.

7KHUH�ZLOO�EH�����PHPEHUV�DIWHU���\HDUV���

Graph each function.����f (x) = |x í 1|

62/87,21���Make a table of values. �

�

x f (x) ±3 4 ±2 3 ±1 2 0 1 1 0 2 1 3 2 4 3

����f (x) = í|2x|

62/87,21���The negative in front of the absolute value makes the graph point down. Make a table of values.�

�

x f (x) ±2 ±4 ±1 ±2 0 0 1 ±2 2 ±4

����f (x) =

62/87,21���Make a table of values. �

�

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

����f (x) =

62/87,21���This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. �

�

x f (x) ±2 ±5 ±1 ±3 0 ±1 1 1 2 ±1 3 0 4 1

����Determine the domain and range of the function graphed below.

62/87,21���The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = �í 2, so the domain is all real numbers and the range is {y | y �í 2}.

eSolutions Manual - Powered by Cognero Page 11

Practice Test - Chapter 9

Use a table of values to graph the following functions. State the domain and range.���y = x2 + 2x + 5

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The graph opens upwards, so the vertex is a minimum located at (í1, 4). So the range is all real numbers greater than or equal to the minimum value of 4, or R = {y | y ����`�

x y �í3 8 í2 5 í1 4 0 5 1 8 2 13

���y = 2x2 í 3x + 1

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The range is all real numbers greater than or equal to the minimum value.

R = {y | y �� í0.125}

x y í2 15 í1 6 0 1 1 0 2 3 3 10

Consider y = x2 í 7x + 6.���Determine whether the function has a maximum or minimum value.

62/87,21���For y = x2 ± 7x + 6, a = 1, b = ±7, and c = 6. Because a is positive, the graph opens upward, so the function has a minimum value.

���State the maximum or minimum value.

62/87,21���Method 1: Find the vertex of the parabola. First, find the x-coordinate of the vertex. For the function y = x2 ± 7x + 6, use a = 1, b = ±7, and c = 6. �

� Substitute 3.5 for x in the function to find the value of the y -FRRUGLQDWH� �

� So, the minimum value is ±6.25. Method 2: Use a graphing calculator to graph the related function f (x) = x2 ± 7x + 6. Use the minimum option on the 2nd [CALC] menu to determine the minimum value.

Therefore, the minimum value is ±6.25.

���What are the domain and range?

62/87,21���Any number can be substituted for x in the function. So, tKH�GRPDLQ�LV�'� �^DOO�UHDO�QXPEHUV`�� The minimum of the function is í6.25. Therefore, R = {y | y �� í6.25} because the range is all real numbers that are JUHDWHU�WKDQ�RU�HTXDO�WR�WKH�PLQLPXP�YDOXH��

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

���x2 + 7x + 10 = 0

62/87,21���Graph the related function f (x) = x2 + 7x + 10.

The x-intercepts of the graph appear to be at ±5 and ±2, so the solutions are ±5 and ±2.Check:

and

���x2 í 5 = í3x

62/87,21���Write the equation in standard form.

Graph the related function f (x) = x2 + 3x í 5.

The x-intercepts are located between ±5 and ±4 and between 1 and 2. Make a table using an increment of 0.1 for the x-values located between ±5 and ±4 and between 1 and 2.

�

For each table, the function value that is closest to zero when the sign changes is 0.04. Thus, the roots are approximately ±4.2 and 1.2.

x ±4.5 ±4.4 ±4.3 ±4.2 ±4.1 y 1.75 1.16 0.59 0.04 ±0.49

x 1.1 1.2 1.3 1.4 1.5 y ±0.49 0.04 0.59 1.16 1.75

Describe how the graph of each function is related to the graph of f (x) = x2.���g(x) = x2 í 5

62/87,21���

The graph of f (x) = x2 + c represents a vertical translation of the parent graph. The value of c is ±5, and ±5 < 0. If c < 0, the graph of f (x) = x2 is translated �XQLWV�GRZQ��7KHUHIRUH��WKH�JUDSK�RI�y = x2 ± 5 is a translation of the graph

of y = x2 shifted down 5 units.

���g(x) = í3x2

62/87,21���

�7KH�JUDSK�RI�f (x) = ax2 expands or compresses the parent graph vertically. Since |a| > 1, the graph is vertically expanded. Since a < 1, the graph is reflected across the x-axis. So, the graph of y = í3x2 is the graph of y = x2 reflected across the x-axis and vertically expanded. �

����

62/87,21���

The function can be written f (x) = ax2 + c, where a = �DQG�c = 4. Since 4 > 0 and 0 < ������WKH�JUDSK�RI�y =

x2 + 4 is the graph of y = x2 vertically compressed and shifted up 4 units.

����08/7,3/(�&+2,&(� Which is an equation for the function shown in the graph?

$� y = í3x2 %� y = 3x2 + 1 &� y = x2 + 2 '� y = í3x2 + 2

62/87,21���The parabola opens downward, so the a in the equation must be negative. Therefore, choices B and C can be eliminated. The y-intercept is at the point (0, c), or in this case (0, 2). Therefore c = 2, so choice A can be eliminated.So, the correct choice is choice D.

Solve each equation by completing the square.����x2 + 2x + 5 = 0

62/87,21���

� No real number has a negative square. So, this equation has no real solutions.

����x2 í x í 6 = 0

62/87,21���

� The solutions are 3 and ±2.

����2x2 í 36 = í6x

62/87,21���Rewrite the equation in standard form.

� The solutions are 3 and ±6.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.����x2 í x í 30 = 0

62/87,21���For this equation, a = 1, b = ±1, and c = ±30.

� The solutions are 6 and ±5.

����x2 í 10x = í15

62/87,21���Rewrite the equation in standard form. �

� For this equation, a = 1, b = ±10, and c = 15.

� The solutions are 1.8 and 8.2.

����2x2 + x ± 15 = 0

62/87,21���For this equation, a = 2, b = 1, and c = ±15.

� The solutions are 2.5 and ±3.

����%$6(%$//��Elias hits a baseball into the air. The equation h = ±16t2 + 60t +3 models the height h in feet of the ball after t seconds. How long is the ball in the air?

62/87,21���

� The ball is in the air for about 3.8 seconds.

����Graph {(í2, 4), (í1, 1), (0, 0), (1, 1), (2, 4)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

62/87,21���

The points are reflected across a line of symmetry. They represent a quadratic function.

����Look for a pattern in the table to determine which kind of model best describes the data.

62/87,21���

� Calculate the first differences.

6LQFH�WKH�ILUVW�GLIIHUHQFHV�DUH�HTXDO��D�OLQHDU�IXQFWLRQ�PRGHOV�WKH�GDWD��Find the slope and y -LQWHUFHSW��

� Use the slope and one point from the table to find the y -LQWHUFHSW��

The data describes the linear function f (x) = 2x + 1.

����CAR CLUB The table shows the number of car club members for four consecutive years after it began. �

� a. Determine which model best represents the data. b. Write a function that models the data. c. Predict the number of car club members after 6 years.

62/87,21���a. First compare the first differences in the number of members:

� Since they're not equal, compare the second differences:

� Since the second differences are not equal, look for a common ratio:

� 7KHUH�LV�D�FRPPRQ�UDWLR�RI����VR�WKH�EHVW�PRGHO�ZRXOG�EH�DQ�H[SRQHQWLDO�IXQFWLRQ�WR�ILW�WKH�GDWD�� � b. A function that fits the data should be an exponential function with a ratio of 2. We are looking for a function of the form . Use a point from the data given to determine the value of a.

� � The function that best fits the data is:

� c. To find the number of car club members after 6 years, use the function found in part b and plug in 6 for x.

7KHUH�ZLOO�EH�����PHPEHUV�DIWHU���\HDUV���

Graph each function.����f (x) = |x í 1|

62/87,21���Make a table of values. �

�

x f (x) ±3 4 ±2 3 ±1 2 0 1 1 0 2 1 3 2 4 3

����f (x) = í|2x|

62/87,21���The negative in front of the absolute value makes the graph point down. Make a table of values.�

�

x f (x) ±2 ±4 ±1 ±2 0 0 1 ±2 2 ±4

����f (x) =

62/87,21���Make a table of values. �

�

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

����f (x) =

62/87,21���This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. �

�

x f (x) ±2 ±5 ±1 ±3 0 ±1 1 1 2 ±1 3 0 4 1

����Determine the domain and range of the function graphed below.

62/87,21���The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = �í 2, so the domain is all real numbers and the range is {y | y �í 2}.

eSolutions Manual - Powered by Cognero Page 12

Practice Test - Chapter 9

Use a table of values to graph the following functions. State the domain and range.���y = x2 + 2x + 5

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The graph opens upwards, so the vertex is a minimum located at (í1, 4). So the range is all real numbers greater than or equal to the minimum value of 4, or R = {y | y ����`�

x y �í3 8 í2 5 í1 4 0 5 1 8 2 13

���y = 2x2 í 3x + 1

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The range is all real numbers greater than or equal to the minimum value.

R = {y | y �� í0.125}

x y í2 15 í1 6 0 1 1 0 2 3 3 10

Consider y = x2 í 7x + 6.���Determine whether the function has a maximum or minimum value.

62/87,21���For y = x2 ± 7x + 6, a = 1, b = ±7, and c = 6. Because a is positive, the graph opens upward, so the function has a minimum value.

���State the maximum or minimum value.

62/87,21���Method 1: Find the vertex of the parabola. First, find the x-coordinate of the vertex. For the function y = x2 ± 7x + 6, use a = 1, b = ±7, and c = 6. �

� Substitute 3.5 for x in the function to find the value of the y -FRRUGLQDWH� �

� So, the minimum value is ±6.25. Method 2: Use a graphing calculator to graph the related function f (x) = x2 ± 7x + 6. Use the minimum option on the 2nd [CALC] menu to determine the minimum value.

Therefore, the minimum value is ±6.25.

���What are the domain and range?

62/87,21���Any number can be substituted for x in the function. So, tKH�GRPDLQ�LV�'� �^DOO�UHDO�QXPEHUV`�� The minimum of the function is í6.25. Therefore, R = {y | y �� í6.25} because the range is all real numbers that are JUHDWHU�WKDQ�RU�HTXDO�WR�WKH�PLQLPXP�YDOXH��

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

���x2 + 7x + 10 = 0

62/87,21���Graph the related function f (x) = x2 + 7x + 10.

The x-intercepts of the graph appear to be at ±5 and ±2, so the solutions are ±5 and ±2.Check:

and

���x2 í 5 = í3x

62/87,21���Write the equation in standard form.

Graph the related function f (x) = x2 + 3x í 5.

The x-intercepts are located between ±5 and ±4 and between 1 and 2. Make a table using an increment of 0.1 for the x-values located between ±5 and ±4 and between 1 and 2.

�

For each table, the function value that is closest to zero when the sign changes is 0.04. Thus, the roots are approximately ±4.2 and 1.2.

x ±4.5 ±4.4 ±4.3 ±4.2 ±4.1 y 1.75 1.16 0.59 0.04 ±0.49

x 1.1 1.2 1.3 1.4 1.5 y ±0.49 0.04 0.59 1.16 1.75

Describe how the graph of each function is related to the graph of f (x) = x2.���g(x) = x2 í 5

62/87,21���

The graph of f (x) = x2 + c represents a vertical translation of the parent graph. The value of c is ±5, and ±5 < 0. If c < 0, the graph of f (x) = x2 is translated �XQLWV�GRZQ��7KHUHIRUH��WKH�JUDSK�RI�y = x2 ± 5 is a translation of the graph

of y = x2 shifted down 5 units.

���g(x) = í3x2

62/87,21���

�7KH�JUDSK�RI�f (x) = ax2 expands or compresses the parent graph vertically. Since |a| > 1, the graph is vertically expanded. Since a < 1, the graph is reflected across the x-axis. So, the graph of y = í3x2 is the graph of y = x2 reflected across the x-axis and vertically expanded. �

����

62/87,21���

The function can be written f (x) = ax2 + c, where a = �DQG�c = 4. Since 4 > 0 and 0 < ������WKH�JUDSK�RI�y =

x2 + 4 is the graph of y = x2 vertically compressed and shifted up 4 units.

����08/7,3/(�&+2,&(� Which is an equation for the function shown in the graph?

$� y = í3x2 %� y = 3x2 + 1 &� y = x2 + 2 '� y = í3x2 + 2

62/87,21���The parabola opens downward, so the a in the equation must be negative. Therefore, choices B and C can be eliminated. The y-intercept is at the point (0, c), or in this case (0, 2). Therefore c = 2, so choice A can be eliminated.So, the correct choice is choice D.

Solve each equation by completing the square.����x2 + 2x + 5 = 0

62/87,21���

� No real number has a negative square. So, this equation has no real solutions.

����x2 í x í 6 = 0

62/87,21���

� The solutions are 3 and ±2.

����2x2 í 36 = í6x

62/87,21���Rewrite the equation in standard form.

� The solutions are 3 and ±6.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.����x2 í x í 30 = 0

62/87,21���For this equation, a = 1, b = ±1, and c = ±30.

� The solutions are 6 and ±5.

����x2 í 10x = í15

62/87,21���Rewrite the equation in standard form. �

� For this equation, a = 1, b = ±10, and c = 15.

� The solutions are 1.8 and 8.2.

����2x2 + x ± 15 = 0

62/87,21���For this equation, a = 2, b = 1, and c = ±15.

� The solutions are 2.5 and ±3.

����%$6(%$//��Elias hits a baseball into the air. The equation h = ±16t2 + 60t +3 models the height h in feet of the ball after t seconds. How long is the ball in the air?

62/87,21���

� The ball is in the air for about 3.8 seconds.

����Graph {(í2, 4), (í1, 1), (0, 0), (1, 1), (2, 4)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

62/87,21���

The points are reflected across a line of symmetry. They represent a quadratic function.

����Look for a pattern in the table to determine which kind of model best describes the data.

62/87,21���

� Calculate the first differences.

6LQFH�WKH�ILUVW�GLIIHUHQFHV�DUH�HTXDO��D�OLQHDU�IXQFWLRQ�PRGHOV�WKH�GDWD��Find the slope and y -LQWHUFHSW��

� Use the slope and one point from the table to find the y -LQWHUFHSW��

The data describes the linear function f (x) = 2x + 1.

����CAR CLUB The table shows the number of car club members for four consecutive years after it began. �

� a. Determine which model best represents the data. b. Write a function that models the data. c. Predict the number of car club members after 6 years.

62/87,21���a. First compare the first differences in the number of members:

� Since they're not equal, compare the second differences:

� Since the second differences are not equal, look for a common ratio:

� 7KHUH�LV�D�FRPPRQ�UDWLR�RI����VR�WKH�EHVW�PRGHO�ZRXOG�EH�DQ�H[SRQHQWLDO�IXQFWLRQ�WR�ILW�WKH�GDWD�� � b. A function that fits the data should be an exponential function with a ratio of 2. We are looking for a function of the form . Use a point from the data given to determine the value of a.

� � The function that best fits the data is:

� c. To find the number of car club members after 6 years, use the function found in part b and plug in 6 for x.

7KHUH�ZLOO�EH�����PHPEHUV�DIWHU���\HDUV���

Graph each function.����f (x) = |x í 1|

62/87,21���Make a table of values. �

�

x f (x) ±3 4 ±2 3 ±1 2 0 1 1 0 2 1 3 2 4 3

����f (x) = í|2x|

62/87,21���The negative in front of the absolute value makes the graph point down. Make a table of values.�

�

x f (x) ±2 ±4 ±1 ±2 0 0 1 ±2 2 ±4

����f (x) =

62/87,21���Make a table of values. �

�

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

����f (x) =

62/87,21���This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. �

�

x f (x) ±2 ±5 ±1 ±3 0 ±1 1 1 2 ±1 3 0 4 1

����Determine the domain and range of the function graphed below.

62/87,21���The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = �í 2, so the domain is all real numbers and the range is {y | y �í 2}.

eSolutions Manual - Powered by Cognero Page 13

Practice Test - Chapter 9

Use a table of values to graph the following functions. State the domain and range.���y = x2 + 2x + 5

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The graph opens upwards, so the vertex is a minimum located at (í1, 4). So the range is all real numbers greater than or equal to the minimum value of 4, or R = {y | y ����`�

x y �í3 8 í2 5 í1 4 0 5 1 8 2 13

���y = 2x2 í 3x + 1

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The range is all real numbers greater than or equal to the minimum value.

R = {y | y �� í0.125}

x y í2 15 í1 6 0 1 1 0 2 3 3 10

Consider y = x2 í 7x + 6.���Determine whether the function has a maximum or minimum value.

62/87,21���For y = x2 ± 7x + 6, a = 1, b = ±7, and c = 6. Because a is positive, the graph opens upward, so the function has a minimum value.

���State the maximum or minimum value.

62/87,21���Method 1: Find the vertex of the parabola. First, find the x-coordinate of the vertex. For the function y = x2 ± 7x + 6, use a = 1, b = ±7, and c = 6. �

� Substitute 3.5 for x in the function to find the value of the y -FRRUGLQDWH� �

� So, the minimum value is ±6.25. Method 2: Use a graphing calculator to graph the related function f (x) = x2 ± 7x + 6. Use the minimum option on the 2nd [CALC] menu to determine the minimum value.

Therefore, the minimum value is ±6.25.

���What are the domain and range?

62/87,21���Any number can be substituted for x in the function. So, tKH�GRPDLQ�LV�'� �^DOO�UHDO�QXPEHUV`�� The minimum of the function is í6.25. Therefore, R = {y | y �� í6.25} because the range is all real numbers that are JUHDWHU�WKDQ�RU�HTXDO�WR�WKH�PLQLPXP�YDOXH��

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

���x2 + 7x + 10 = 0

62/87,21���Graph the related function f (x) = x2 + 7x + 10.

The x-intercepts of the graph appear to be at ±5 and ±2, so the solutions are ±5 and ±2.Check:

and

���x2 í 5 = í3x

62/87,21���Write the equation in standard form.

Graph the related function f (x) = x2 + 3x í 5.

The x-intercepts are located between ±5 and ±4 and between 1 and 2. Make a table using an increment of 0.1 for the x-values located between ±5 and ±4 and between 1 and 2.

�

For each table, the function value that is closest to zero when the sign changes is 0.04. Thus, the roots are approximately ±4.2 and 1.2.

x ±4.5 ±4.4 ±4.3 ±4.2 ±4.1 y 1.75 1.16 0.59 0.04 ±0.49

x 1.1 1.2 1.3 1.4 1.5 y ±0.49 0.04 0.59 1.16 1.75

Describe how the graph of each function is related to the graph of f (x) = x2.���g(x) = x2 í 5

62/87,21���

The graph of f (x) = x2 + c represents a vertical translation of the parent graph. The value of c is ±5, and ±5 < 0. If c < 0, the graph of f (x) = x2 is translated �XQLWV�GRZQ��7KHUHIRUH��WKH�JUDSK�RI�y = x2 ± 5 is a translation of the graph

of y = x2 shifted down 5 units.

���g(x) = í3x2

62/87,21���

�7KH�JUDSK�RI�f (x) = ax2 expands or compresses the parent graph vertically. Since |a| > 1, the graph is vertically expanded. Since a < 1, the graph is reflected across the x-axis. So, the graph of y = í3x2 is the graph of y = x2 reflected across the x-axis and vertically expanded. �

����

62/87,21���

The function can be written f (x) = ax2 + c, where a = �DQG�c = 4. Since 4 > 0 and 0 < ������WKH�JUDSK�RI�y =

x2 + 4 is the graph of y = x2 vertically compressed and shifted up 4 units.

����08/7,3/(�&+2,&(� Which is an equation for the function shown in the graph?

$� y = í3x2 %� y = 3x2 + 1 &� y = x2 + 2 '� y = í3x2 + 2

62/87,21���The parabola opens downward, so the a in the equation must be negative. Therefore, choices B and C can be eliminated. The y-intercept is at the point (0, c), or in this case (0, 2). Therefore c = 2, so choice A can be eliminated.So, the correct choice is choice D.

Solve each equation by completing the square.����x2 + 2x + 5 = 0

62/87,21���

� No real number has a negative square. So, this equation has no real solutions.

����x2 í x í 6 = 0

62/87,21���

� The solutions are 3 and ±2.

����2x2 í 36 = í6x

62/87,21���Rewrite the equation in standard form.

� The solutions are 3 and ±6.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.����x2 í x í 30 = 0

62/87,21���For this equation, a = 1, b = ±1, and c = ±30.

� The solutions are 6 and ±5.

����x2 í 10x = í15

62/87,21���Rewrite the equation in standard form. �

� For this equation, a = 1, b = ±10, and c = 15.

� The solutions are 1.8 and 8.2.

����2x2 + x ± 15 = 0

62/87,21���For this equation, a = 2, b = 1, and c = ±15.

� The solutions are 2.5 and ±3.

����%$6(%$//��Elias hits a baseball into the air. The equation h = ±16t2 + 60t +3 models the height h in feet of the ball after t seconds. How long is the ball in the air?

62/87,21���

� The ball is in the air for about 3.8 seconds.

����Graph {(í2, 4), (í1, 1), (0, 0), (1, 1), (2, 4)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

62/87,21���

The points are reflected across a line of symmetry. They represent a quadratic function.

����Look for a pattern in the table to determine which kind of model best describes the data.

62/87,21���

� Calculate the first differences.

6LQFH�WKH�ILUVW�GLIIHUHQFHV�DUH�HTXDO��D�OLQHDU�IXQFWLRQ�PRGHOV�WKH�GDWD��Find the slope and y -LQWHUFHSW��

� Use the slope and one point from the table to find the y -LQWHUFHSW��

The data describes the linear function f (x) = 2x + 1.

����CAR CLUB The table shows the number of car club members for four consecutive years after it began. �

� a. Determine which model best represents the data. b. Write a function that models the data. c. Predict the number of car club members after 6 years.

62/87,21���a. First compare the first differences in the number of members:

� Since they're not equal, compare the second differences:

� Since the second differences are not equal, look for a common ratio:

� 7KHUH�LV�D�FRPPRQ�UDWLR�RI����VR�WKH�EHVW�PRGHO�ZRXOG�EH�DQ�H[SRQHQWLDO�IXQFWLRQ�WR�ILW�WKH�GDWD�� � b. A function that fits the data should be an exponential function with a ratio of 2. We are looking for a function of the form . Use a point from the data given to determine the value of a.

� � The function that best fits the data is:

� c. To find the number of car club members after 6 years, use the function found in part b and plug in 6 for x.

7KHUH�ZLOO�EH�����PHPEHUV�DIWHU���\HDUV���

Graph each function.����f (x) = |x í 1|

62/87,21���Make a table of values. �

�

x f (x) ±3 4 ±2 3 ±1 2 0 1 1 0 2 1 3 2 4 3

����f (x) = í|2x|

62/87,21���The negative in front of the absolute value makes the graph point down. Make a table of values.�

�

x f (x) ±2 ±4 ±1 ±2 0 0 1 ±2 2 ±4

����f (x) =

62/87,21���Make a table of values. �

�

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

����f (x) =

62/87,21���This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. �

�

x f (x) ±2 ±5 ±1 ±3 0 ±1 1 1 2 ±1 3 0 4 1

����Determine the domain and range of the function graphed below.

62/87,21���The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = �í 2, so the domain is all real numbers and the range is {y | y �í 2}.

eSolutions Manual - Powered by Cognero Page 14

Practice Test - Chapter 9

Use a table of values to graph the following functions. State the domain and range.���y = x2 + 2x + 5

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The graph opens upwards, so the vertex is a minimum located at (í1, 4). So the range is all real numbers greater than or equal to the minimum value of 4, or R = {y | y ����`�

x y �í3 8 í2 5 í1 4 0 5 1 8 2 13

���y = 2x2 í 3x + 1

62/87,21���

Graph the ordered pairs, and connect them to create a smooth curve. The parabola extends to infinity.

The function is a parabola, so the domain is all real numbers. The range is all real numbers greater than or equal to the minimum value.

R = {y | y �� í0.125}

x y í2 15 í1 6 0 1 1 0 2 3 3 10

Consider y = x2 í 7x + 6.���Determine whether the function has a maximum or minimum value.

62/87,21���For y = x2 ± 7x + 6, a = 1, b = ±7, and c = 6. Because a is positive, the graph opens upward, so the function has a minimum value.

���State the maximum or minimum value.

62/87,21���Method 1: Find the vertex of the parabola. First, find the x-coordinate of the vertex. For the function y = x2 ± 7x + 6, use a = 1, b = ±7, and c = 6. �

� Substitute 3.5 for x in the function to find the value of the y -FRRUGLQDWH� �

� So, the minimum value is ±6.25. Method 2: Use a graphing calculator to graph the related function f (x) = x2 ± 7x + 6. Use the minimum option on the 2nd [CALC] menu to determine the minimum value.

Therefore, the minimum value is ±6.25.

���What are the domain and range?

62/87,21���Any number can be substituted for x in the function. So, tKH�GRPDLQ�LV�'� �^DOO�UHDO�QXPEHUV`�� The minimum of the function is í6.25. Therefore, R = {y | y �� í6.25} because the range is all real numbers that are JUHDWHU�WKDQ�RU�HTXDO�WR�WKH�PLQLPXP�YDOXH��

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

���x2 + 7x + 10 = 0

62/87,21���Graph the related function f (x) = x2 + 7x + 10.

The x-intercepts of the graph appear to be at ±5 and ±2, so the solutions are ±5 and ±2.Check:

and

���x2 í 5 = í3x

62/87,21���Write the equation in standard form.

Graph the related function f (x) = x2 + 3x í 5.

The x-intercepts are located between ±5 and ±4 and between 1 and 2. Make a table using an increment of 0.1 for the x-values located between ±5 and ±4 and between 1 and 2.

�

For each table, the function value that is closest to zero when the sign changes is 0.04. Thus, the roots are approximately ±4.2 and 1.2.

x ±4.5 ±4.4 ±4.3 ±4.2 ±4.1 y 1.75 1.16 0.59 0.04 ±0.49

x 1.1 1.2 1.3 1.4 1.5 y ±0.49 0.04 0.59 1.16 1.75

Describe how the graph of each function is related to the graph of f (x) = x2.���g(x) = x2 í 5

62/87,21���

The graph of f (x) = x2 + c represents a vertical translation of the parent graph. The value of c is ±5, and ±5 < 0. If c < 0, the graph of f (x) = x2 is translated �XQLWV�GRZQ��7KHUHIRUH��WKH�JUDSK�RI�y = x2 ± 5 is a translation of the graph

of y = x2 shifted down 5 units.

���g(x) = í3x2

62/87,21���

�7KH�JUDSK�RI�f (x) = ax2 expands or compresses the parent graph vertically. Since |a| > 1, the graph is vertically expanded. Since a < 1, the graph is reflected across the x-axis. So, the graph of y = í3x2 is the graph of y = x2 reflected across the x-axis and vertically expanded. �

����

62/87,21���

The function can be written f (x) = ax2 + c, where a = �DQG�c = 4. Since 4 > 0 and 0 < ������WKH�JUDSK�RI�y =

x2 + 4 is the graph of y = x2 vertically compressed and shifted up 4 units.

����08/7,3/(�&+2,&(� Which is an equation for the function shown in the graph?

$� y = í3x2 %� y = 3x2 + 1 &� y = x2 + 2 '� y = í3x2 + 2

62/87,21���The parabola opens downward, so the a in the equation must be negative. Therefore, choices B and C can be eliminated. The y-intercept is at the point (0, c), or in this case (0, 2). Therefore c = 2, so choice A can be eliminated.So, the correct choice is choice D.

Solve each equation by completing the square.����x2 + 2x + 5 = 0

62/87,21���

� No real number has a negative square. So, this equation has no real solutions.

����x2 í x í 6 = 0

62/87,21���

� The solutions are 3 and ±2.

����2x2 í 36 = í6x

62/87,21���Rewrite the equation in standard form.

� The solutions are 3 and ±6.

Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.����x2 í x í 30 = 0

62/87,21���For this equation, a = 1, b = ±1, and c = ±30.

� The solutions are 6 and ±5.

����x2 í 10x = í15

62/87,21���Rewrite the equation in standard form. �

� For this equation, a = 1, b = ±10, and c = 15.

� The solutions are 1.8 and 8.2.

����2x2 + x ± 15 = 0

62/87,21���For this equation, a = 2, b = 1, and c = ±15.

� The solutions are 2.5 and ±3.

����%$6(%$//��Elias hits a baseball into the air. The equation h = ±16t2 + 60t +3 models the height h in feet of the ball after t seconds. How long is the ball in the air?

62/87,21���

� The ball is in the air for about 3.8 seconds.

����Graph {(í2, 4), (í1, 1), (0, 0), (1, 1), (2, 4)}. Determine whether the ordered pairs represent a linear function, a quadratic function, or an exponential function.

62/87,21���

The points are reflected across a line of symmetry. They represent a quadratic function.

����Look for a pattern in the table to determine which kind of model best describes the data.

62/87,21���

� Calculate the first differences.

6LQFH�WKH�ILUVW�GLIIHUHQFHV�DUH�HTXDO��D�OLQHDU�IXQFWLRQ�PRGHOV�WKH�GDWD��Find the slope and y -LQWHUFHSW��

� Use the slope and one point from the table to find the y -LQWHUFHSW��

The data describes the linear function f (x) = 2x + 1.

����CAR CLUB The table shows the number of car club members for four consecutive years after it began. �

� a. Determine which model best represents the data. b. Write a function that models the data. c. Predict the number of car club members after 6 years.

62/87,21���a. First compare the first differences in the number of members:

� Since they're not equal, compare the second differences:

� Since the second differences are not equal, look for a common ratio:

� 7KHUH�LV�D�FRPPRQ�UDWLR�RI����VR�WKH�EHVW�PRGHO�ZRXOG�EH�DQ�H[SRQHQWLDO�IXQFWLRQ�WR�ILW�WKH�GDWD�� � b. A function that fits the data should be an exponential function with a ratio of 2. We are looking for a function of the form . Use a point from the data given to determine the value of a.

� � The function that best fits the data is:

� c. To find the number of car club members after 6 years, use the function found in part b and plug in 6 for x.

7KHUH�ZLOO�EH�����PHPEHUV�DIWHU���\HDUV���

Graph each function.����f (x) = |x í 1|

62/87,21���Make a table of values. �

�

x f (x) ±3 4 ±2 3 ±1 2 0 1 1 0 2 1 3 2 4 3

����f (x) = í|2x|

62/87,21���The negative in front of the absolute value makes the graph point down. Make a table of values.�

�

x f (x) ±2 ±4 ±1 ±2 0 0 1 ±2 2 ±4

����f (x) =

62/87,21���Make a table of values. �

�

x f (x) 0 0

0.5 0 1 1

1.5 1 2 2

2.5 2 3 3

����f (x) =

62/87,21���This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the function changes. �

�

x f (x) ±2 ±5 ±1 ±3 0 ±1 1 1 2 ±1 3 0 4 1

����Determine the domain and range of the function graphed below.

62/87,21���The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = �í 2, so the domain is all real numbers and the range is {y | y �í 2}.

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Practice Test - Chapter 9