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§ 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.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 2.4
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).
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 2.4
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).
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 2.4
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
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 2.4
Graphing Using Intercepts 128
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)
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 2.4
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
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 2.4
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.
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 2.4
Slopes of Lines 130
Check Point 2 bCheck Point 2 b
SOLUTIONSOLUTION
Find the slope of the line passing through the pair of points. Then explain what the slope means.
.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:
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 2.4
Slopes of Lines 130
Possibilities for a Line’s SlopePositive Slope Negative Slope
Zero Slope Undefined Slope
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 2.4
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
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 2.4
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.
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 2.4
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.
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 2.4
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
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 2.4
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.
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 2.4
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)
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 2.4
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
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 2.4
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)
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 2.4
Graphing Lines 133
Check Point 4Check Point 4
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.
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 2.4
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)
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 2.4
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)
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 2.4
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.
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 2.4
Slope as Rate of Change p 136
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
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 2.4
Slope as Rate of Change p 136
Check Point 8Check Point 8
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
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 2.4
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.
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 2.4
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.
DONE
Blitzer, Intermediate Algebra, 5e – Slide #28 Section 2.4
Graphing Using Intercepts 129
3) Find a checkpoint, a third ordered-pair solution. Let x=4.
Check Point 1Check Point 1
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
Blitzer, Intermediate Algebra, 5e – Slide #29 Section 2.4
Slopes of Lines
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
Find the slope of the line passing through the pair of points. Then explain what the slope means.
.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.
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 2.4
Slope as Rate of Change p 136
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.
Blitzer, Intermediate Algebra, 5e – Slide #31 Section 2.4
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.
Blitzer, Intermediate Algebra, 5e – Slide #32 Section 2.4
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.