§ 2.4

§ 2.4

Linear Functions and Slope

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 2.4

x- and y-Intercepts 127

Using Intercepts to Graph Ax + By = C1) Find the x-intercept. Let y = 0 and solve for x.

2) Find the y-intercept. Let x = 0 and solve for y.

3) Find a checkpoint, a third ordered-pair solution.

4) Graph the equation by drawing a line through the three points.

Ax + By = C is called the standard form of the equation of the line.


Graphing Using Intercepts 128

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Graph the equation using intercepts.

1242 yx

1) Find the x-intercept. Let y = 0 and then solve for x.

1242 yx

12042 x

122 x

6x

Replace y with 0

Multiply and simplify

Divide by -2

The x-intercept is -6, so the line passes through (-6,0).


Graphing Using Intercepts

1242 yx

2) Find the y-intercept. Let x = 0 and then solve for y.

12402 y

124 y

3y

Replace x with 0

Multiply and simplify

Divide by 4

CONTINUECONTINUEDD

The y-intercept is 3, so the line passes through (0,3).


Graphing Using Intercepts

3) Find a checkpoint, a third ordered-pair solution. For our checkpoint, we will let x = 1 (because x = 1 is not the x-intercept) and find the corresponding value for y.

12412 y

1242 y

144 y

Replace x with 1

Multiply

Add 2 to both sides

CONTINUECONTINUEDD

The checkpoint is the ordered pair (1,3.5).

1242 yx

5.32

7

4

14y Divide by 4 and simplify



4) Graph the equation by drawing a line through the three points. The three points in the figure below lie along the same line. Drawing a line through the three points results in the graph of .

CONTINUECONTINUEDD

1242 yx

(1,3.5)

(-6,0)

(0,3)


Slope of a Line 129

The slope m of the line through the two distinct points

12

12

in Change

in Change

xx

yy

x

ym

),( 11 yx and ),( 22 yx is m, where

Note that you may call either point However, don’t subtract in one order in the numerator and use another order for subtraction in the denominator - or the sign of your slope will be wrong.

),( 11 yx


Slopes of Lines 130

Check Point 2 aCheck Point 2 a

SOLUTIONSOLUTION

Find the slope of the line passing through the pair of points. Then explain what the slope means.

.2,4, and 4,3,Let 2211 yxyx

6

1

6

34

)4(2

34

42

in Change

in Change

12

12

xx

yy

x

ym

(-3,4) and (-4,-2)

The slope is obtained as follows:

This means that for every six units that the line (that passes through both points) travels up, it also travels 1 units to the right.


Slopes of Lines 130

Check Point 2 bCheck Point 2 b

SOLUTIONSOLUTION


.5,1, and 2,4,Let 2211 yxyx

5

7

5

7

)4(1

)2(5

41

25

in Change

in Change

12

12

xx

yy

x

ym

(4,-2) and (-1,5)

The slope is obtained as follows:


Slopes of Lines 130

Possibilities for a Line’s SlopePositive Slope Negative Slope

Zero Slope Undefined Slope


Slopes of Lines 131

Slope-Intercept Form of the Equation of a Line

y = mx + b with slope m, and y-intercept b

Examplesy = .968x – 17.3 m = .968; b = -17.3

f (x) = 3-5x m = -5; b = 3


Slopes of Lines 131

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Give the slope and y-intercept for the line whose equation is:

5

73

5-

5-

xy

5

7

5

3 xy

3x - 5y = 7

Subtract 3x from both sides3x - 3x - 5y = -3x+7-5y = -3x+7 Simplify

Divide both sides by -5

Simplify

Therefore, the slope is 3/5 and the y-intercept is -7/5.

3x - 5y = 7Hint: You should solve for y so as to have the equation in slope intercept form.


Slopes of Lines 132

Check Point 3Check Point 3

SOLUTIONSOLUTION

Give the slope and y-intercept for the line whose equation is:

4

208

4-

4-

xy

52 xy

8x - 4y = 20

Subtract 8x from both sides8x - 8x - 4y = -8x+20

-4y = -8x+20 Simplify

Divide both sides by -4

Simplify

Therefore, the slope is 2 and the y-intercept is -5.

8x - 4y = 20Solve for y so as to have the equation in slope intercept form.


Slopes of Lines 132

Graphing y = mx + b Using the Slope and y-Intercept1) Plot the point containing the y-intercept on the y-axis. This is the point (0,b).

2) Obtain a second point using the slope, m. Write m as a fraction, and use rise-over-run, starting at the point containing the y-intercept, to plot this point.

3) Use a straightedge to draw a line through the two points. Draw arrowheads at the ends of the line to show that the line continues indefinitely in both directions


Graphing Lines 132

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Graph the line whose equation is f (x) = 2x + 3.

The equation f (x) = 2x + 3 is in the form y = mx + b.

Now that we have identified the slope and the y-intercept, we use the three steps in the box to graph the equation.

The slope is 2 and the y-intercept is 3.


Graphing Lines

1) Plot the point containing the y-intercept on the y-axis. The y-intercept is 3. We plot the point (0,3), shown below.

CONTINUECONTINUEDD

(0,3)


Graphing Lines

2) Obtain a second point using the slope, m. Write m as a fraction, and use rise over run, starting at the point containing the y-intercept, to plot this point. We express the slope, 2, as a fraction.

CONTINUECONTINUEDD

(0,3)We plot the second point on the line by starting at (0,3), the first point. Based on the slope, we move 2 units up (the rise) and 1 unit to the right (the run). This puts us at a second point on the line, (1,-1), shown on the graph.

(1,5)Run

Rise

1

2m


Graphing Lines

3) Use a straightedge to draw a line through the two points. The graph of y = 2x + 3 is show below.

CONTINUECONTINUEDD

(0,3)(1,5)


Graphing Lines 133


SOLUTIONSOLUTION

Graph the line whose equation is f (x) = 4x - 3.

The equation f (x) = 4x - 3 is in the form y = mx + b.

With the slope and the y-intercept, graph the equation.

The slope is 4 and the y-intercept is -3.


Horizontal and Vertical Lines 134

EXAMPLE 6 on page 134EXAMPLE 6 on page 134

SOLUTIONSOLUTION

Graph the linear equation: y = -4.

All ordered pairs that are solutions of y = -4 have a value of y that is always -4. Any value can be used for x. Let’s select three of the possible values for x: -3, 1, 6:

x y = -4 (x,y)

-3 -4 (-3,-4)

1 -4 (1,-4)

6 -4 (6,-4)

Upon plotting the three resultant points and connecting the points with a line, the graph to the right is the solution.

(1,-4)

(-3,-4) (6,-4)


Horizontal and Vertical Lines 135

EXAMPLE – similar to Ex 7 on page 135EXAMPLE – similar to Ex 7 on page 135

SOLUTIONSOLUTION

Graph the linear equation: x = 5.

All ordered pairs that are solutions of x = 5 have a value of x that is always 5. Any value can be used for y. Let’s select three of the possible values for y: -3, 1, 5:

x = 5 y (x,y)

5 -3 (5,-3)

5 1 (5,1)

5 5 (5,5)

Upon plotting the three resultant points and connecting the points with a line, the graph to the right is the solution.

(5,1)

(5,-3)

(5,5)


Slope as Rate of Change p 136

A linear function that models data is described. Slope may be interpreted as the rate of change of the dependent variable per unit change in the independent variable.

•Find the slope of each model.

•Then describe what this means in terms of the rate of change of the dependent variable y per unit change in the independent variable x.



EXAMPLE EXAMPLE 88Using the Figure 2.27 graph on page 136 which show the percentage of

Americans in two age groups who reported using illegal drugs in the previous month.

SOLUTIONSOLUTION

• Find the slope of the line for the 12-17 age group.

• The slope indicates that for the 12-17 age group, the percentage reporting using illegal drugs in the previous month decreased by approximately 0.57% per year. The rate of change is a decrease of approximately 0.57%.

57.03

7.1

20022005

6.119.9

12

12

xx

yym




Using the Figure 2.27 graph on page 136 to find the slope of the line segment for the 50-59 age group. Express the slope correct to two decimal places and describes what it represents.

SOLUTIONSOLUTION

• Find the slope of the line for the 50-59 age group.

• For the 50-67 age group, the percentage reporting using illegal drugs in the previous month increased by approximately 0.57% per year.

57.03

7.1

20022005

7.24.4

m


Modeling with Slope-Intercept p 139

EXAMPLE 10EXAMPLE 10

Using the Figure 2.31(b) scatter plot on page 138 find a function in the form T(x)=mx+b.that models the average ticket price for a movie T(x), x years after 1990 and use the model to predict the average ticket price in 2010.

SOLUTIONSOLUTION

• Find the slope.

15.0015

23.441.6

12

12

xx

yym

23.415.0)( xxT

• The average ticket price for a movie, T(x), x years after 1990 can be modeled by the linear function. The slope indicates that ticket prices increase $0.15 per year.


Modeling with Slope-Intercept p 139

EXAMPLE 10, EXAMPLE 10, continuedcontinued

23.723.4)20(15.0)20( T

• Because 2010 is 20 years after 1990, substitute 20 for x and evaluate.

• The model predicts an average ticket price of $7.23 in 2010.

• Try doing Check Point 10 on your own.

Note that vertical intercept and y-intercept are the same.



3) Find a checkpoint, a third ordered-pair solution. Let x=4.


The checkpoint is the ordered pair (4,3).

Graph the equation using intercepts. 623 yx

1) Find the x-intercept. Let y = 0 and then solve for x.

2) Find the y-intercept. Let x = 0 and then solve for y.

The y-intercept is -3, so the line passes through (0,-3).

The x-intercept is 2, so the line passes through (2,0).

2 63 6023 xxx

3y 62y- 6203 y

3y 62y- 6243 y


Slopes of Lines

EXAMPLEEXAMPLE

SOLUTIONSOLUTION


.2,4, and 8,7,Let 2211 yxyx

11

6

74

82

74

82

in Change

in Change

12

12

xx

yy

x

ym

(-7,-8) and (4,-2)

Remember, the slope is obtained as follows:

This means that for every six units that the line (that passes through both points) travels up, it also travels 11 units to the right.



EXAMPLEEXAMPLE

The linear function y = 2x + 24 models the average cost in dollars of a retail drug prescription in the United States, y, x years after 1991.

SOLUTIONSOLUTION

• The slope of the linear model is 2.

• This means that every year (since 1991) the average cost in dollars of a retail drug prescription in the U.S. has increased approximately $2.


Slope as Rate of Change 130

The change in the y values is the vertical change or the “Rise”.

The change in the x values is the horizontal change or the “Run.”

m is commonly used to denote the slope of a line. The absolute value of a line is related to its steepness.

When the slope is positive, the line rises from left to right.

When the slope is negative, the line falls from left to right.


Slope Intercept Form 131

In the form of the equation of the line where the equation is written as y = mx + b, it is true that if x = 0, then y = b.

Therefore, b is the y intercept for the equation of the line. Furthermore, the x coefficient m is the slope of the line.

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