1. Rate of Change and Slope
ALGEBRA 1 LESSON 6-1
(For help, go to Lessons 5-2 and 1-5.)
Evaluate each function rule for x = –5.
1. y = x – y = 7 – x
3. y = 2x y = – x + 3
Write in simplest form. 2 5
7 – 3
3 – 1
3 – 5
6 – 0
8 – (–4)
3 – 7
–1 – 2
0 – 5
–6 – (–4)
–2 – 6
0 – 1
1 – 0
6-1

2. Rate of Change and Slope
ALGEBRA 1 LESSON 6-1
Solutions
1. y = x – 7 for x = –5: y = –5 – 7 = –12
2. y = 7 – x for x = –5: y = 7 – (–5) = 12
3. y = 2x + 5 for x = –5: y = 2(–5) + 5 = – = –5
4. y = – x + 3 for x = –5: y = – (–5) + 3 = = 5
5. = = = – = –
7. = – = – = =
= = = – = –1 2 5 2 5
7 – 3
3 – 1 4 2
3 – 5
6 – 0 2 6 1 3
8 – (–4)
3 – 7 12 4
–1 – 2
0 – 5 –3 –5 3 5
–6 – (–4)
–2 – 6 –2 –8 1 4
0 – 1
1 – 0 1
6-1

3. Rate of Change and Slope
ALGEBRA 1 LESSON 6-1
For the data in the table, is the rate of change the same for each pair of consecutive mileage amounts?
Fee for Miles Driven
Miles
Fee
100
$30
150
200
250
$42
$54
$66
Find the rate of change for each pair
of consecutive mileage amounts.
6-1

4. Rate of Change and Slope
ALGEBRA 1 LESSON 6-1
(continued)
change in cost Cost depends on the
change in number of miles number of miles.
rate of change =
42 – – –
150– – – =
The rate of change for each pair of consecutive mileage amounts is $12 per 50 miles. The rate of change is the same for all the data.
6-1

5. Rate of Change and Slope
ALGEBRA 1 LESSON 6-1
Below is a graph of the distance traveled by a motorcycle from its starting point. Find the rate of change. Explain what this rate of change means.
Choose two points
On the graph (0,0) and
(400,20)
6-1

6. Rate of Change and Slope
ALGEBRA 1 LESSON 6-1
(continued)
Using the points (0,0) and (400,20), find the rate of change.
vertical change change in distance
horizontal change change in time
rate of change =
400 – 0
20 – 0
400 20 =
Use two points.
Divide the vertical change by
the horizontal change.
Simplify.
The rate of change is 20 m/s.
The motorcycle is traveling 20 meters each second.
6-1

7. Rate of Change and Slope
ALGEBRA 1 LESSON 6-1
Find the slope of each line.
slope =
rise
run a.
= – 3 2 = –2
4 – 1
0 – 2
The slope of the line is – . 3 2
6-1

8. Rate of Change and Slope
ALGEBRA 1 LESSON 6-1
(continued)
slope =
rise
run b.
Find the slope of the line.
= 2 = –2 –1
–1 – 1
–2 – (–2)
The slope of the line is 2.
6-1

9. Rate of Change and Slope
ALGEBRA 1 LESSON 6-1
Find the slope of each line through E(3, –2) and F(–2, –1).
slope =
y2 – y1
x2 – x1
= – 1 5 = –5
Substitute (–2, –1) for (x2, y2) and
(3, –2) for (x1, y1).
Simplify.
–1 – (–2)
–2 – 3
The slope of EF is – . 1 5
6-1

10. Rate of Change and Slope
ALGEBRA 1 LESSON 6-1
Find the slope of each line. a.
slope =
y2 – y1
x2 – x1
2 – 2
1 – (–4) = 5
Substitute (1, 2) for (x2, y2) and (–4, 2) for (x1, y1).
Simplify.
= 0
The slope of the horizontal line is 0.
6-1

11. Rate of Change and Slope
ALGEBRA 1 LESSON 6-1
(continued)
slope =
y2 – y1
x2 – x1 b.
Find the slope of the line. = 5
Substitute (2, 1) for (x2, y2) and (2, –4) for (x1, y1).
1 – (–4)
2 – 2
Simplify.
Division by zero is undefined.
The slope of the vertical line in undefined.
6-1

12. Rate of Change and Slope
ALGEBRA 1 LESSON 6-1
1. Find the rate of change
for the data in the table. 5
$75 6 7 8
$90
$105
$120
Tickets
Cost
2. Find the rate of change
for the data in the graph.
3. Find the slope of the line.
4. Find the slope of the line through (3, –2) and (–2, 5).
6-1

13. Rate of Change and Slope
ALGEBRA 1 LESSON 6-1
5. State whether the slope is zero or undefined. a. b.
6-1

14. Rate of Change and Slope
ALGEBRA 1 LESSON 6-1
1. Find the rate of change
for the data in the table.
2. Find the rate of change
for the data in the graph.
3. Find the slope of the line. 5
$75 6 7 8
$90
$105
$120
Tickets
Cost
$15 per ticket 4 3
400 calories per hour 7 5 –
4. Find the slope of the line through (3, –2) and (–2, 5).
6-1

15. Rate of Change and Slope
ALGEBRA 1 LESSON 6-1
5. State whether the slope is zero or undefined. a. b.
undefined
6-1

16. Evaluate each expression. 1. 6a + 3 for a = 2 2. –2x – 5 for x = 3
Slope-Intercept Form
ALGEBRA 1 LESSON 6-2
(For help, go to Lessons 1-6 and 2-6.)
Evaluate each expression.
1. 6a + 3 for a = –2x – 5 for x = 3
3. x + 2 for x = x + 2 for x = 15
Solve each equation for y.
5. y – 5 = 4x 6. y + 2x = 7
7. 2y + 6 = –8x 1 4
6-2

17. Slope-Intercept Form Solutions
ALGEBRA 1 LESSON 6-2
Solutions
1. 6a + 3 for a = 2: 6(2) + 3 = = 15
2. –2x – 5 for x = 3: –2(3) – 5 = –6 – 5 = –11
3. x + 2 for x = 16: (16) + 2 = = 6
4. 0.2x + 2 for x = 15: 0.2(15) + 2 = = 5
y – 5 = 4x y + 2x = 7
y – = 4x y + 2x – 2x = 7 – 2x
y = 4x y = –2x + 7
y + 6 = –8x
2y + 6 – 6 = –8x – 6
2y = –8x – 6 =
y = –4x – 3 1 4 1 4 2y y
–8x – 6 2
6-2

18. What are the slope and y-intercept of y = 2x – 3?
Slope-Intercept Form
ALGEBRA 1 LESSON 6-2
What are the slope and y-intercept of y = 2x – 3?
mx + b =
2x – 3
Use the Slope-Intercept form.
The slope is 2; the y-intercept is –3.
6-2

19. Write an equation of the line with slope and y-intercept 4.
Slope-Intercept Form
ALGEBRA 1 LESSON 6-2
Write an equation of the line with slope and
y-intercept 4. 2 5 2 5
Use the slope-intercept form.
Substitute for m and 4 for b.
y = mx + b
y = x + 4
6-2

20. Write an equation for the line.
Slope-Intercept Form
ALGEBRA 1 LESSON 6-2
Write an equation for the line.
Step 1 Find the slope. Two points on the line are (0, 1) and (3, –1).
–1 – 1
3 – 0
slope =
= – 2 3 2 3
y = mx + b
y = – x + 1
Substitute – for m and 1 for b.
Step 2 Write an equation in slope–
intercept form. The y-intercept is 1.
6-2

21. The slope is . Use slope to plot a second point. Step 3
Slope-Intercept Form
ALGEBRA 1 LESSON 6-2
Graph y = x – 2. 1 3
Step 1
The y-intercept is –2.
So plot a point at (0,–2).
Step 2
The slope is . Use slope to plot a second point. 1 3
Step 3
Draw a line through
the two points.
6-2

22. The base pay for a used car salesperson is $300 per week.
Slope-Intercept Form
ALGEBRA 1 LESSON 6-2
The base pay for a used car salesperson is $300 per week.
The salesperson also earns 15% commission on sales made. The equation t = s relates total earnings t to sales s. Graph the equations.
Step 1  Identify the slope and y-intercept.
t = s
t = 0.15s + 300  Rewrite the equation in slope intercept form.
slope y-intercept
6-2

23. Step 3 Draw a line through the points.
Slope-Intercept Form
ALGEBRA 1 LESSON 6-2
(continued)
Step 2  Plot two points. First plot a point at the y-intercept. Then use the slope to plot a second point. The slope is 0.15, which equals Plot a second point 15 units above and 100 units to the right of the y-intercept. 15
100
Step 3  Draw a line through the points.
6-2

24. Find the slope and y-intercept of each equation.
Slope-Intercept Form
ALGEBRA 1 LESSON 6-2
Find the slope and y-intercept of each equation.
1. y = x – 3
2. y = –2x
Write an equation of a line with the given slope and y-intercept.
3. m = – , b = –4
4. m = 0.5, b = 1 3 4 5
5. Write an equation for the line.
m = ; b = –3 3 4
y = x – 3 1 2
m = –2; b = 0
y = – x – 4 3 5
6. Graph y = –x + 3.
y = 0.5x + 1
6-2

25. Solve each equation for y. 1. 3x + y = 5 2. y – 2x = 10 3. x – y = 6
Standard Form
ALGEBRA 1 LESSON 6-3
(For help, go to Lessons 2-3 and 2-6.)
Solve each equation for y.
1. 3x + y = y – 2x = x – y = 6
4. 20x + 4y = y + 3x = y – 2x = 4
Clear each equation of decimals.
x = = 0.2x – 5
– 0.222x = 1
6-3

26. Standard Form Solutions 1. 3x + y = 5 2. y – 2x = 10
ALGEBRA 1 LESSON 6-3
Solutions
x + y = y – 2x = 10
3x – 3x + y = 5 – 3x y – 2x + 2x = x
y = –3x y = 2x + 10
3. x – y = x + 4y = 8
x = 6 + y y = –20x + 8
x – 6 = y y =
y = x – y = –5x + 2
5. 9y + 3x = y – 2x = 4
9y = –3x y = 2x + 4
y = y =
y = – x y = x +
7. Multiply each term by 100:
100(6.25x)+100(8.5) = 100(7.75)
Simplify:
625x = 775
8. Multiply each term by 10:
10(0.4) = 10(0.2x) – 10(5)
4 = 2x – 50
9. Multiply each term by 1000:
1000(0.9)–1000(0.222x)=1000(1)
900 – 222x = 1000
–20x + 8 4
–3x+ 1 9
2x+ 4 5 1 3 1 9 2 5 4 5
6-3

27. Find the x- and y-intercepts of 2x + 5y = 6.
Standard Form
ALGEBRA 1 LESSON 6-3
Find the x- and y-intercepts of 2x + 5y = 6.
Step 1  To find the x-intercept, substitute 0 for y and
solve for x.
2x + 5y = x + 5(0) = x = x = 3
The x-intercept is 3.
Step 2  To find the y-intercept,
substitute 0 for x and
solve for y.
2x + 5y = 6
2(0) + 5y = 6
5y = 6
y =
The y-intercept is . 6 5
6-3

28. Graph 3x + 5y = 15 using intercepts.
Standard Form
ALGEBRA 1 LESSON 6-3
Graph 3x + 5y = 15 using intercepts.
Step 1  Find the intercepts.
3x + 5y = 15
3x + 5(0) = 15 Substitute 0 for y.
3x = 15 Solve for x.
x = 5
3(0) + 5y = 15 Substitute 0 for x.
5y = 15 Solve for y.
y = 3
Step 2  Plot (5, 0) and (0, 3).
Draw a line through the points.
6-3

29. 0 • x + 1 • y = 4 Write in standard form. For all values of x, y = 4.
ALGEBRA 1 LESSON 6-3
a. Graph y = 4
b. Graph x = –3.
0 • x + 1 • y = 4 Write in standard form.
For all values of x, y = 4.
1 • x + 0 • y = –3 Write in standard form.
For all values of y, x = –3.
6-3

30. Write y = x + 6 in standard form using integers.
ALGEBRA 1 LESSON 6-3
Write y = x + 6 in standard form using integers. 2 3
y = x + 6 2 3
3y = 3( x + 6 ) Multiply each side by 3.
3y = 2x Use the Distributive Property.
–2x + 3y = Subtract 2x from each side.
The equation in standard form is –2x + 3y = 18.
6-3

31. Define: Let = the hours mowing lawns.
Standard Form
ALGEBRA 1 LESSON 6-3
Write an equation in standard form to find the number of hours you would need to work at each job to make a total of $130.
Define:  Let = the hours mowing lawns.
Let = the hours delivering newspapers. x y
Amount Paid per hour
Job
Mowing
lawns
$12
Delivering
newspapers
Relate: $12 per h   plus $5 per h equals   $130
mowing delivering $5
Write:  = x y
The equation standard form is 12x + 5y = 130.
6-3

32. Find each x- and y-intercepts of each equation.
Standard Form
ALGEBRA 1 LESSON 6-3
Find each x- and y-intercepts of each equation.
1. 3x + y = –4x – 3y = 9
3. Graph 2x – y = 6 using x- and y-intercepts.
For each equation, tell whether its graph is horizontal or vertical.
4. y = x = –8
6. Write y = x – 3 in standard form using integers.
x = – , y = –3 9 4
x = 4, y = 12
horizontal
vertical 5 2
–5x + 2y = –6 or 5x – 2y = 6
6-3

33. Point-Slope Form and Writing Linear Equations
ALGEBRA 1 LESSON 6-4
(For help, go to Lessons 6-1 and 1-7.)
Find the rate of change of the data in each table.
Simplify each expression.
4. –3(x – 5) 5. 5(x + 2) 6. – (x – 6) 4 9 2 5 8 11 –2 –8
–14 x y –3 –5 –1 1 3 –4 10
7.5
2.5 –6
–11
6-4

34. Point-Slope Form and Writing Linear Equations
ALGEBRA 1 LESSON 6-4
1. Use points (2, 4) and (5, –2). rate of change = = = –2
2. Use points (–3, –5) and (–1, –4). rate of change = = = =
3. Use points (10, 4) and (7.5, –1). rate of change = = = = 2
4. –3(x – 5) = –3x – (–3)(5) = –3x + 15
5. 5(x + 2) = 5x + 5(2) = 5x + 10
6. – (x – 6) = – x – (– )(6) = – x + 4 9 8 3
–5 + 4
–3 + 1
4 – (–2)
2 – 5
4 – (–1 )
10 – 7.5
–5 – (–4)
–3 – (–1)
4 + 1
2.5 5 –1 –2 6 –3 1 2
Solutions
6-4

35. Point-Slope Form and Writing Linear Equations
ALGEBRA 1 LESSON 6-4 1 3
Graph the equation y – 2 = (x – 1). 1 3
The equation of a line that passes through (1, 2) with slope .
Start at (1, 2). Using the slope, go up 1 unit and right 3 units to (4, 3).
Draw a line through the two points.
6-4

36. Point-Slope Form and Writing Linear Equations
ALGEBRA 1 LESSON 6-4
Write the equation of the line with slope –2 that passes through the point (3, –3).
y – y1 = m(x – x1)
Substitute (3, –3) for (x1, y1) and –2 for m.
y – (–3) = –2(x – 3)
Simplify the grouping symbols.
y + 3 = –2(x – 3)
6-4

37. Point-Slope Form and Writing Linear Equations
ALGEBRA 1 LESSON 6-4
Write equations for the line in point-slope form and in
slope-intercept form.
The slope is – . 1 3
Step 1  Find the slope.
= m
y2 – y1
x2 – x1
4 – 3
–1 – 2
= –
6-4

38. Point-Slope Form and Writing Linear Equations
ALGEBRA 1 LESSON 6-4
(continued)
Step 2  Use either point to write the the equation in point-slope form.
Use (–1, 4).
y – y1 = m(x – x1)
y – 4 = – (x + 1) 1 3
Step 3  Rewrite the equation
from Step 2 in slope–
intercept form.
y – 4 = – (x + 1)
y – 4 = – x –
y = – x + 3 1 3 2
6-4

39. Point-Slope Form and Writing Linear Equations
ALGEBRA 1 LESSON 6-4
Is the relationship shown by the data linear? If so, model the data with an equation.
Step 1  Find the rate of change for
consecutive ordered pairs. –2 –1 –6 –3
= 2 3 6 2 –1 –3 4 –2 –6 x y
–1( ) –2
–3( ) –6
–2( ) –4
The relationship is linear.
The rate of change is 2.
6-4

40. Point-Slope Form and Writing Linear Equations
ALGEBRA 1 LESSON 6-4
(continued)
Step 2 Use the slope 2 and a point (2,4) to write an equation.
y – y1 = m(x – x1) Use the point-slope form.
y – 4 = 2(x – 2) Substitute (2, 4) for (x1, y1) and 2 for m.
6-4

41. Point-Slope Form and Writing Linear Equations
ALGEBRA 1 LESSON 6-4
Is the relationship shown by the data linear? If so, model the data with an equation.
Step 1  Find the rate of change for
consecutive ordered pairs. 1 2
= 1 – / –2 –1 1 2 x y
1 ( ) 1
2 ( ) 1
The relationship is not linear.
6-4

42. Point-Slope Form and Writing Linear Equations
ALGEBRA 1 LESSON 6-4
1. Graph the equation y + 1 = –(x – 3).
2. Write an equation of the line with
slope – that passes through
the point (0, 4).
3. Write an equation for the line that
passes through (3, –5) and (–2, 1) in
Point-Slope form and Slope-Intercept
form.
4. Is the relationship shown by the data
linear? If so, model that data with an
equation. 2 3
y – 4 = – (x – 0), or y = – x + 4 2 3
–10 –7 5 20 –3 –1 x y 6 5 7
y + 5 = – (x – 3); y = – x – 2 5
yes; y + 3 = (x – 0)
6-4

43. Parallel and Perpendicular Lines
ALGEBRA 1 LESSON 6-5
(For help, go to Lessons 1-6 and 6-2.)
What is the reciprocal of each fraction?
– 4. –
What are the slope and y-intercept of each equation?
5. y = x y = x – 8 7. y = 6x 8. y = 6x – 2 1 2 4 3 2 5 7 5 5 3 5 3
6-5

44. Parallel and Perpendicular Lines
ALGEBRA 1 LESSON 6-5
1. The reciprocal of is or The reciprocal of is .
3. The reciprocal of – is – The reciprocal of – is – .
5. y = x y = x – 8
slope: slope:
y-intercept: y-intercept: –8
7. y = 6x y = 6x – 2
slope: slope: 6
y-intercept: y-intercept: – 2 1 2 4 3 5 7
Solutions
6-5

45. Parallel and Perpendicular Lines
ALGEBRA 1 LESSON 6-5
Are the graphs of y = –2x – 1 and 4x + 2y = 6 parallel?
Write 4x + 2y = 6 in slope-intercept form. Then compare with y = –2x – 1.
2y = –4x + 6 Subtract 4x from each side.
y = –2x + 3 Simplify.
Divide each side by 2.
2y –4x + 6 =
The lines are parallel. The equations have the same slope, –2, and different y-intercepts.
6-5

46. Parallel and Perpendicular Lines
ALGEBRA 1 LESSON 6-5
Write an equation for the line that contains (–2, 3) and is parallel to y = x – 4. 5 2
Step 1 Identify the slope
of the given line.
y = x – 4
slope 5 2
Step 2  Write the equation of the line through (–2, 3)
using slope-intercept form.
y – y1 = m(x – x1) Use point-slope form.
y – 3 = (x + 2) Substitute (–2, 3) for (x1, y1) and for m. 5 2
y – 3 = x + (2) Use the Distributive Property. 5 2
y – 3 = x Simplify. 5 2
y = x Add 3 to each side
and simplify. 5 2
6-5

47. Parallel and Perpendicular Lines
ALGEBRA 1 LESSON 6-5
Write an equation of the line that contains (6, 2) and is perpendicular to y = –2x + 7.
Step 1 Identify the slope of the given line.
y = – 2 x + 7
slope
Step 2  Find the negative reciprocal of the slope.
The negative reciprocal of –2 is . 1 2
6-5

48. Parallel and Perpendicular Lines
ALGEBRA 1 LESSON 6-5
(continued)
Step 3  Use the slope-intercept form to write an equation.
y = mx + b
2 = (6) + b Substitute for m, 6 for x, and 2 for y. 1 2 1 2
2 = 3 + b Simplify.
2 – 3 = 3 + b – Subtract 3 from each side.
–1 = b Simplify.
The equation is y = x – 1. 1 2
6-5

49. Parallel and Perpendicular Lines
ALGEBRA 1 LESSON 6-5
The line in the graph represents the street in front of a new house. The point is the front door. The sidewalk from the front door will be perpendicular to the street. Write an equation representing the sidewalk.
Step 1  Find the slope m of the street.
m = = = = – Points (0, 2) and (3, 0) are on the street.
y2 – y1
x2 – x1
0 – 2
3 – 0 –2 3 2
6-5

50. Parallel and Perpendicular Lines
ALGEBRA 1 LESSON 6-5
(continued)
Step 2 Find the negative reciprocal of the slope.
The negative reciprocal of – is . So the slope of the sidewalk is . The y-intercept is –3. 2 3
The equation for the
sidewalk is y = x – 3. 3 2
6-5

51. Parallel and Perpendicular Lines
ALGEBRA 1 LESSON 6-5 3 2
1. Find the slope of a line parallel to 3x – 2y = 1.
2. Find the slope of a line perpendicular to 4x + 5y = 7.
Tell whether the lines for each pair of equations are parallel, perpendicular, or neither.
3. y = 3x – 1, y = – x + 2
4. y = 2x + 5, 2x + y = –4
5. y = – x – 1, 2x + 3y = 6
6. Write an equation of the line that contains (2, 1) and is
perpendicular to y = – x + 3. 5 4 1 3
perpendicular
neither 2 3
parallel 1 2
y = 2x – 3
6-5

52. Scatter Plots and Equations of Lines
ALGEBRA 1 LESSON 6-6
1. Graph the data in a scatter plot.
Draw a trend line.
2. Write an equation for the trend
line.
3. Predict the number of households
in the U.S. in (Use 105 for x.)
4. Use a calculator to find the line of
best fit for the data.
5. What is the correlation coefficient?
Number of Households in the U.S.
[Source: U.S. Census Bureau, Current Population Reports.
From Statistical Abstract of the United States, 2000]
1980
80.8
1985
1990
1995
86.8
93.3
99.0
Year
Households
(millions)
1975
71.1
y – 86.8 = 1.3(x – 85)
about million households
y = 1.366x – 29.91
6-6

53. Graphing Absolute Value Equations
ALGEBRA 1 LESSON 6-7
1. Describe how the graphs of y = |x| and y = |x| – 9 are the same
and how they are different.
2. Graph y = |x| – 2 by translating y = |x|.
3. Write an equation for the translation of y = |x|, 1.5 units down.
4. Graph y = |x – 5|.
5. Write an equation for the translation of y = |x|, 7 units left.
The graphs are the same shape. The y-intercept of the first graph is 0.
The y-intercept of the second graph is –9.
y = |x| – 1.5
y = |x + 7|
6-7