5 3 the graphs of quadratic equations-x

Graphs of Quadratic Equations

The graphs of 1st degree equations Ax + By = C are straight lines hence they are called linear equations.

Graphs of Quadratic Equations

The graphs of 1st degree equations Ax + By = C are straight lines hence they are called linear equations. If we have a linear function in the form of y = f(x) = mx + b, then we may eyeball the graph, starting from the y-intercept b,

Graphs of Quadratic Equations

For example, for y = -2x + 3, we start at the point (0, 3).

the graph of y = -2x + 3

The graphs of 1st degree equations Ax + By = C are straight lines hence they are called linear equations. If we have a linear function in the form of y = f(x) = mx + b, then we may eyeball the graph, starting from the y-intercept b,

Graphs of Quadratic Equations

For example, for y = -2x + 3, we start at the point (0, 3).

The graphs of 1st degree equations Ax + By = C are straight lines hence they are called linear equations. If we have a linear function in the form of y = f(x) = mx + b, then we may eyeball the graph, starting from the y-intercept b, eyeball the tilt using the slope m.

Graphs of Quadratic Equations

the graph of y = -2x + 3

For example, for y = -2x + 3, we start at the point (0, 3). The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1,

The graphs of 1st degree equations Ax + By = C are straight lines hence they are called linear equations. If we have a linear function in the form of y = f(x) = mx + b, then we may eyeball the graph, starting from the y-intercept b, eyeball the tilt using the slope m.

Graphs of Quadratic Equations

the graph of y = -2x + 3

For example, for y = -2x + 3, we start at the point (0, 3). The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1, i.e. another point is located 2 down and 1 to the right from (0, 3)

The graphs of 1st degree equations Ax + By = C are straight lines hence they are called linear equations. If we have a linear function in the form of y = f(x) = mx + b, then we may eyeball the graph, starting from the y-intercept b, eyeball the tilt using the slope m.

Graphs of Quadratic Equations

the graph of y = -2x + 3

(0, 3)

∆y = -2

the graph of y = -2x + 3

The graphs of 1st degree equations Ax + By = C are straight lines hence they are called linear equations. If we have a linear function in the form of y = f(x) = mx + b, then we may eyeball the graph, starting from the y-intercept b, eyeball the tilt using the slope m.

Graphs of Quadratic Equations

For example, for y = -2x + 3, we start at the point (0, 3). The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1, i.e. another point is located 2 down and 1 to the right from (0, 3)

(0, 3)

(1, 1)∆y = -2 ∆x = 1

the graph of y = -2x + 3

The graphs of 1st degree equations Ax + By = C are straight lines hence they are called linear equations. If we have a linear function in the form of y = f(x) = mx + b, then we may eyeball the graph, starting from the y-intercept b, eyeball the tilt using the slope m.

Graphs of Quadratic Equations

For example, for y = -2x + 3, we start at the point (0, 3). The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1, i.e. another point is located 2 down and 1 to the right from (0, 3)

(0, 3)

(1, 1)∆y = -2 ∆x = 1

the graph of y = -2x + 3

The graphs of 1st degree equations Ax + By = C are straight lines hence they are called linear equations. If we have a linear function in the form of y = f(x) = mx + b, then we may eyeball the graph, starting from the y-intercept b, eyeball the tilt using the slope m.

Graphs of Quadratic Equations

For example, for y = -2x + 3, we start at the point (0, 3). The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1, i.e. another point is located 2 down and 1 to the right from (0, 3) and the line contains both points.

For example, for y = -2x + 3, we start at the point (0, 3). The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1, i.e. another point is located 2 down and 1 to the right from (0, 3) and the line contains both points.

The graphs of 2nd degree equations y = ax2 + bx + c are not straight. They are curves. We will develop a method for eyeballing the graphs of y = ax2 + bx + c in this section.

(0, 3)

(1, 1)∆y = -2 ∆x = 1

the graph of y = -2x + 3

The graphs of 1st degree equations Ax + By = C are straight lines hence they are called linear equations. If we have a linear function in the form of y = f(x) = mx + b, then we may eyeball the graph, starting from the y-intercept b, eyeball the tilt using the slope m.

Graphs of Quadratic Equations

Graphs of Quadratic EquationsWe start with an example of a graph gives the general shape of the graphs of 2nd (quadratic) degree equations.

Graphs of Quadratic EquationsWe start with an example of a graph gives the general shape of the graphs of 2nd (quadratic) degree equations.

Example A. Graph y = –x2

Graphs of Quadratic EquationsWe start with an example of a graph gives the general shape of the graphs of 2nd (quadratic) degree equations.

Make a tableExample A. Graph y = –x2

x -4 -3 -2 -1 0 1 2 3 4

Graphs of Quadratic EquationsWe start with an example of a graph gives the general shape of the graphs of 2nd (quadratic) degree equations.

Make a tableExample A. Graph y = –x2

y

x -4 -3 -2 -1 0 1 2 3 4

Graphs of Quadratic EquationsWe start with an example of a graph gives the general shape of the graphs of 2nd (quadratic) degree equations.

Make a tableExample A. Graph y = –x2

y-16-9-4-1 0-1-4-9-16

x -4 -3 -2 -1 0 1 2 3 4

Graphs of Quadratic EquationsWe start with an example of a graph gives the general shape of the graphs of 2nd (quadratic) degree equations.

Make a tableExample A. Graph y = –x2

y-16-9-4-1 0-1-4-9-16

Graphs of Quadratic EquationsWe start with an example of a graph gives the general shape of the graphs of 2nd (quadratic) degree equations.

Make a tableExample A. Graph y = –x2

x -4 -3 -2 -1 0 1 2 3 4

y-16-9-4-1 0-1-4-9-16

The graphs of 2nd (quadratic) equations are called parabolas. Graphs of Quadratic Equations

The graphs of 2nd (quadratic) equations are called parabolas. Parabolas describe the paths of thrown objects (or the upside-down paths).

Graphs of Quadratic Equations

The graphs of 2nd (quadratic) equations are called parabolas. Parabolas describe the paths of thrown objects (or the upside-down paths).

Graphs of Quadratic Equations

Properties of Parabolas:

The graphs of 2nd (quadratic) equations are called parabolas. Parabolas describe the paths of thrown objects (or the upside-down paths).

Graphs of Quadratic Equations

Properties of Parabolas: • Parabolas are symmetric with respect to a center line

The graphs of 2nd (quadratic) equations are called parabolas. Parabolas describe the paths of thrown objects (or the upside-down paths).

Graphs of Quadratic Equations

Properties of Parabolas: • Parabolas are symmetric with respect to a center line• The highest or lowest point of the parabola sits on the center line. This point is called the vertex, or the tip.

The graphs of 2nd (quadratic) equations are called parabolas. Parabolas describe the paths of thrown objects (or the upside-down paths).

Graphs of Quadratic Equations

Properties of Parabolas: • Parabolas are symmetric with respect to a center line• The highest or lowest point of the parabola sits on the center line. This point is called the vertex, or the tip.

the vertex

the vertex

The graphs of 2nd (quadratic) equations are called parabolas. Parabolas describe the paths of thrown objects (or the upside-down paths).

Graphs of Quadratic Equations

Properties of Parabolas: • Parabolas are symmetric with respect to a center line• The highest or lowest point of the parabola sits on the center line. This point is called the vertex.The vertex is the starting point for graphing a parabola, i.e. the graph of y = ax2 + bx + c, a 2nd degree function.

the vertex

the vertex

The graphs of 2nd (quadratic) equations are called parabolas. Parabolas describe the paths of thrown objects (or the upside-down paths).

Graphs of Quadratic Equations

Properties of Parabolas: • Parabolas are symmetric with respect to a center line• The highest or lowest point of the parabola sits on the center line. This point is called the vertex.

Vertex Formula: The vertex of y = ax2 + bx + c is at x = . –b2a

The vertex is the starting point for graphing a parabola, i.e. the graph of y = ax2 + bx + c, a 2nd degree function.

the vertex

the vertex

Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.

Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.

Example B. Graph y = x2 – 4x – 12

Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.

Example B. Graph y = x2 – 4x – 12

Vertex: set x = = 2 –(–4) 2(1)

Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.

Example B. Graph y = x2 – 4x – 12

Vertex: set x = = 2 –(–4) 2(1)

Make a table centered at x = 2. x y 0 1 2 3 4

Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.

Example B. Graph y = x2 – 4x – 12

Vertex: set x = = 2 –(–4) 2(1)

Make a table centered at x = 2. x y 0 1 2 3 4

–12 –15 –16–15–12

Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.

Example B. Graph y = x2 – 4x – 12

Vertex: set x = = 2 –(–4) 2(1)

Make a table centered at x = 2. x y 0 1 2 3 4

–12 –15 –16–15–12

Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.

Example B. Graph y = x2 – 4x – 12

Vertex: set x = = 2 –(–4) 2(1)

Make a table centered at x = 2. x y 0 1 2 3 4 Note the y values are symmetric

around the vertex just as the points.

–12 –15 –16–15–12

Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.

Example B. Graph y = x2 – 4x – 12

Vertex: set x = = 2 –(–4) 2(1)

Make a table centered at x = 2. x y 0 1 2 3 4 Note the y values are symmetric

around the vertex just as the points. (If they are not, check your calculation.)

–12 –15 –16–15–12

Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.

Example B. Graph y = x2 – 4x – 12

Vertex: set x = = 2 –(–4) 2(1)

Make a table centered at x = 2. x y 0 1 2 3 4 Note the y values are symmetric

around the vertex just as the points. (If they are not, check your calculation.)

(2, -16)

–12 –15 –16–15–12

Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.

Example B. Graph y = x2 – 4x – 12

Vertex: set x = = 2 –(–4) 2(1)

Make a table centered at x = 2. x y 0 1 2 3 4 Note the y values are symmetric

around the vertex just as the points. (If they are not, check your calculation.)

(2, -16)(3, -15)

–12 –15 –16–15–12

Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.

Example B. Graph y = x2 – 4x – 12

Vertex: set x = = 2 –(–4) 2(1)

Make a table centered at x = 2. x y 0 1 2 3 4 Note the y values are symmetric

around the vertex just as the points. (If they are not, check your calculation.)

(2, -16)

(4, -12)

(3, -15)

–12 –15 –16–15–12

Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.

Example B. Graph y = x2 – 4x – 12

Vertex: set x = = 2 –(–4) 2(1)

Make a table centered at x = 2. x y 0 1 2 3 4 Note the y values are symmetric

around the vertex just as the points. (If they are not, check your calculation.)

(2, -16)

(4, -12)

(3, -15)(1, -15)

–12 –15 –16–15–12

Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.

Example B. Graph y = x2 – 4x – 12

Vertex: set x = = 2 –(–4) 2(1)

Make a table centered at x = 2. x y 0 1 2 3 4 Note the y values are symmetric

around the vertex just as the points. (If they are not, check your calculation.)

(2, -16)

(0, -12) (4, -12)

(3, -15)(1, -15)

–12 –15 –16–15–12

Graphs of Quadratic Equations

Example B. Graph y = x2 – 4x – 12

Vertex: set x = = 2 –(–4) 2(1)

(2, -16)

(0, -12) (4, -12)

Make a table centered at x = 2.

Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.)

(3, -15)(1, -15)

One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.

x y 0 1 2 3 4

–12 –15 –16–15–12

Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-intercepts and the y-intercept.

Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-intercepts and the y-intercept. The y-intercept is (0, c) obtained by setting x = 0.

Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-intercepts and the y-intercept. The y-intercept is (0, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c

Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-intercepts and the y-intercept. The y-intercept is (0, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c which may or may not have real number solutions.

Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-intercepts and the y-intercept. The y-intercept is (0, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.

The center line is determined by the vertex.

Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-intercepts and the y-intercept. The y-intercept is (0, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.

The center line is determined by the vertex. Suppose we know another point on the parabola,

Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-intercepts and the y-intercept. The y-intercept is (0, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.

The center line is determined by the vertex. Suppose we know another point on the parabola, the reflection of the point across the center is also on the parabola.

Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-intercepts and the y-intercept. The y-intercept is (0, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.

The center line is determined by the vertex. Suppose we know another point on the parabola, the reflection of the point across the center is also on the parabola. There is exactly one parabola that goes through these three points.

Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-intercepts and the y-intercept. The y-intercept is (0, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.

(2nd way) To graph a parabola y = ax2 + bx + c.Graphs of Quadratic Equations

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.-b2a

Graphs of Quadratic Equations

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.

-b2a

Graphs of Quadratic Equations

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.3. Locate its reflection across the center line. These three

points form the tip of the parabola.

-b2a

Graphs of Quadratic Equations

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.3. Locate its reflection across the center line. These three

points form the tip of the parabola. Trace the parabola.

-b2a

Graphs of Quadratic Equations

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.3. Locate its reflection across the center line. These three

points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.

-b2a

Graphs of Quadratic Equations

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.3. Locate its reflection across the center line. These three

points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.

-b2a

Graphs of Quadratic Equations

Example C. Graph y = –x2 + 2x + 15

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.3. Locate its reflection across the center line. These three

points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.

-b2a

Graphs of Quadratic Equations

The vertex is at x = 1, y = 16Example C. Graph y = –x2 + 2x + 15

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.3. Locate its reflection across the center line. These three

points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.

-b2a

Graphs of Quadratic Equations

The vertex is at x = 1, y = 16y-intercept is at (0, 15)

Example C. Graph y = –x2 + 2x + 15

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.3. Locate its reflection across the center line. These three

points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.

-b2a

Graphs of Quadratic Equations

The vertex is at x = 1, y = 16y-intercept is at (0, 15) Plot its reflection (2, 15)

Example C. Graph y = –x2 + 2x + 15

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.3. Locate its reflection across the center line. These three

points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.

-b2a

Graphs of Quadratic Equations

The vertex is at x = 1, y = 16y-intercept is at (0, 15) Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0

Example C. Graph y = –x2 + 2x + 15

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.3. Locate its reflection across the center line. These three

points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.

-b2a

Graphs of Quadratic Equations

The vertex is at x = 1, y = 16y-intercept is at (0, 15) Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0

Example C. Graph y = –x2 + 2x + 15

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.3. Locate its reflection across the center line. These three

points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.

-b2a

Graphs of Quadratic Equations

The vertex is at x = 1, y = 16y-intercept is at (0, 15) Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0(x + 3)(x – 5) = 0x = –3, x = 5

Example C. Graph y = –x2 + 2x + 15

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.3. Locate its reflection across the center line. These three

points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.

-b2a

Graphs of Quadratic Equations

The vertex is at x = 1, y = 16y-intercept is at (0, 15) Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0(x + 3)(x – 5) = 0x = –3, x = 5

Example C. Graph y = –x2 + 2x + 15 (1, 16)

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.3. Locate its reflection across the center line. These three

points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.

-b2a

Graphs of Quadratic Equations

The vertex is at x = 1, y = 16y-intercept is at (0, 15) Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0(x + 3)(x – 5) = 0x = –3, x = 5

Example C. Graph y = –x2 + 2x + 15 (1, 16)(0, 15)

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.3. Locate its reflection across the center line. These three

points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.

-b2a

Graphs of Quadratic Equations

The vertex is at x = 1, y = 16y-intercept is at (0, 15) Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0(x + 3)(x – 5) = 0x = –3, x = 5

Example C. Graph y = –x2 + 2x + 15 (1, 16)(0, 15) (2, 15)

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.3. Locate its reflection across the center line. These three

points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.

-b2a

Graphs of Quadratic Equations

The vertex is at x = 1, y = 16y-intercept is at (0, 15) Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0(x + 3)(x – 5) = 0x = –3, x = 5

Example C. Graph y = –x2 + 2x + 15 (1, 16)(0, 15) (2, 15)

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.3. Locate its reflection across the center line. These three

points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.

-b2a

Graphs of Quadratic Equations

The vertex is at x = 1, y = 16y-intercept is at (0, 15) Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0(x + 3)(x – 5) = 0x = –3, x = 5

Example C. Graph y = –x2 + 2x + 15 (1, 16)(0, 15) (2, 15)

(-3, 0) (5, 0)

Finally, we make the observation that given y = ax2 + …,if a > 0, then the parabola opens upward.

Graphs of Quadratic Equations

Finally, we make the observation that given y = ax2 + …,if a > 0, then the parabola opens upward.

Graphs of Quadratic Equations

if a < 0, then the parabola opens downward.

Exercise A. 1. Practice drawing the following parabolas with paper and pencil. Visualize them as the paths of thrown objects. Make sure pay attention to the symmetry.

Graphs of Quadratic Equations

Exercise B. Graph the parabolas by taking a table around the vertex that reflect the symmetry. Find the x and y intercepts.

Graphs of Quadratic Equations

4. y = x2 – 4 5. y = –x2 + 42. y = –x2 3. y = x2

6. y = x2 + 4 7. y = –x2 – 48. y = x2 – 2x – 3 9. y = –x2 + 2x + 310. y = x2 + 2x – 3 11. y = –x2 – 2x + 312. y = x2 – 2x – 8 13. y = –x2 + 2x + 814. y = x2 + 2x – 8 15. y = –x2 – 2x + 816. a. y = x2 – 4x – 5 b. y = –x2 + 4x + 517. a. y = x2 + 4x – 5 b. y = –x2 – 4x + 519. y = x2 + 4x – 21 20. y = x2 – 4x – 45 21. y = x2 – 6x – 27 22. y = – x2 – 6x + 27

Exercise C. Graph the following parabolas by plotting the vertex point, the y–intercept and its reflection. Find the x intercepts.

Graphs of Quadratic Equations

23. y = x2 – 2x – 3 24. y = –x2 + 2x + 3

25. y = x2 + 2x – 3 26. y = –x2 – 2x + 3

27. y = x2 – 2x – 8 28. y = –x2 + 2x + 8

29. y = x2 + 2x – 8 30. y = –x2 – 2x + 8

31. y = x2 + 4x – 21 32. y = x2 – 4x – 45

33. y = x2 – 6x – 27 34. y = – x2 – 6x + 27Exercise D. Graph the following parabolas by plotting the vertex point, the y–intercept and its reflection. Verify that there is no x intercepts (i.e. they have complex roots). 35. y = x2 – 2x + 8 36. y = –x2 + 2x – 537. y = x2 + 2x + 3 38. y = –x2 – 3x – 439. y = 2x2 + 3x + 4 40. y = x2 – 4x + 32

Graphs of Quadratic EquationsAnswers to odd problems.

5. y = –x2 + 4

(0,0)

(0, 4)

(2, 0)

7. y = –x2 – 4

(0, –4)

(2, 0)(–4 , 0)

(–1, 9)

(–2, 0)

15. y = –x2 – 2x + 8 17. a. y = x2 + 4x – 5 b. y = –x2 – 4x + 5

19. y = x2 + 4x – 21 21. y = x2 – 6x – 27

(1, 0)(–5, 0)

(–2 , –9)(0, –5)

(3, 0)(–7, 0)

(–2 , –25)(0, –21)

(1, 0)(–5, 0)

(–2 , 9)

(0, 5)

(9, 0)(–3, 0)

(0 , –27) (0, –21)

(3 , –36)

23. y = x2 – 2x – 3(3, 0)(–1, 0)

(–1 , –4)

(0, –3)

3. y = x2