lesson 3 mi/vocab slope-intercept form write and graph linear equations in slope-intercept form....

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slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope-intercept form.

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Page 1: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

• slope-intercept form

• Write and graph linear equations in slope-intercept form.

• Model real-world data with an equation in slope-intercept form.

Page 3: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

Slope-intercept form

Answer:

Write an equation in slope-intercept form of the line

whose slope is and whose y-intercept is –6.

Write an Equation Given Slope and y-Intercept

Page 4: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

A. y = 3x + 4

B. y = 4x + 3

C. y = 4x

D. y = 4

Write an equation in slope-intercept form of the line whose slope is 4 and whose y-intercept is 3.

Page 5: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

Write an Equation From a Graph

Write an equation in slope-intercept form of the line shown in the graph.

Step 1 You know the coordinates of two points on the line. Find the slope. Let (x1, y1) = (0, –3) and (x2, y2) = (2, 1).

(x1, y1) = (0, –3)

(x2, y2) = (2, 1)

Page 6: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

Write an Equation From a Graph

Simplify.

Answer: The equation of the lines is y = 2x – 3.

The slope is 2.

Step 2 The line crosses the y-axis at (0, –3).So, the y-intercept is –3.

Step 3 Finally, write the equation.

Slope-intercept form

Simplify.

y = mx + b

y = 2x – 3

Page 7: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

1. A

2. B

3. C

4. D

0%0%0%0%

A B C D

Write an equation in slope intercept form of the line shown in the graph.

A.

B.

C.

D.

Page 8: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

Graph Equations

A. Graph y = 0.5x – 7.

Step 1 The y-intercept is –7. So graph (0, –7).

From (0, –7), move up 1 unit and right 2 units. Draw a dot.

Step 3 Draw a line through the points.

Step 2 The slope is 0.5

Page 9: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

Graph Equations

B. Graph 5x + 4y = 8.

Step 1 Solve for y to write the equation in slope-intercept form.

8 – 5x = 8 + (–5x) or –5x + 8

Subtract 5x from each side.

Simplify.

Original equation

Divide each side by 4.

5x + 4y = 8

5x + 4y – 5x = 8 – 5x

4y = 8 – 5x

4y = –5x + 8

Page 10: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

Graph Equations

Divide each term in the numerator by 4.

Step 2 The y-intercept of

So graph (0, 2).

Page 11: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

Graph Equations

From (0, 2), move down 5 units and right 4 units. Draw a dot.

Step 3 The slope is

Step 4 Draw a line connecting the points.

Answer:

Page 12: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

1. A

2. B

3. C

4. D

0%0%0%0%

A B C D

A. Graph y = 2x – 4.

A. B.

C. D.

Page 13: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

1. A

2. B

3. C

4. D

0%0%0%0%

A B C D

B. Graph 3x + 2y = 6.

A. B.

C. D.

Page 14: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

A. HEALTH The ideal maximum heart rate for a 25-year-old who is exercising to burn fat is 117 beats per minute. For every 5 years older than 25, that ideal rate drops 3 beats per minute. Write a linear equation to find the ideal maximum heart rate for anyone over 25 who is exercising to burn fat.

Write an Equation in Slope-Intercept Form

Page 15: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

Answer:

Write an Equation in Slope-Intercept Form

Words

Variables

Equation

Let R = the ideal heart rate.

Let a = years older than 25.

ideal rateIdeal rate of years older for 25- rate equals change times than 25 plus year-old.

R = × a + 117

Page 16: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

B. Graph the equation.

Answer:

Write an Equation in Slope-Intercept Form

The graph passes through (0, 117) with a slope of

Page 17: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

C. Find the ideal maximum heart rate for a person exercising to burn fat who is 55 years old.

Answer: The ideal heart rate for a 55-year-old person is 99 beats per minute.

Write an Equation in Slope-Intercept Form

The age 55 is 30 years older than 25. So, a = 30.

Ideal heart rate equation

Replace a with 30.

Simplify.

Page 18: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

A. D = 0.15n

B. D = 0.15n + 3

C. D = 3n

D. D = 3n + 0.15

A. The amount of money spent on Christmas gifts has increased by an average of $150,000 ($0.15 million) per year since 1986. Consumers spent $3 million in 1986. Write a linear equation to find the average amount spent for any year since 1986.

Page 19: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

A. A

B. B

C. C

D. D

A B C D

0% 0%0%0%

B. The amount of money spent on Christmas gifts has increased by an average of $150,000 ($0.15 million) per year since 1986. Consumers spent $3 million in 1986. Graph the equation.

A.B.

C.D.

Page 20: Lesson 3 MI/Vocab slope-intercept form Write and graph linear equations in slope-intercept form. Model real-world data with an equation in slope- intercept

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

A. $5 million

B. $3 million

C. $4.95 million

D. $3.5 million

C. The amount of money spent on Christmas gifts has increased by an average of $150,000 ($0.15 million) per year since 1986. Consumers spent $3 million in 1986. Find the amount spent by consumers in 1999.