Linear Equations Graphs and Functions. Copyright © 2008 Pearson Addison-Wesley. All rights...

Linear Equations Graphs and Functions

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-2

Linear Functions

Graphing Linear Functions ▪ Standard From Ax + By = C ▪ Slope ▪ Average Rate of Change ▪ Linear Models

Linear Functions


A linear function has the form:

f(x) = mx + b

where m and b are real numbers and m is the slope which comes from the equation

for the given points (x1,y1) and (x2,y2) .

12

12

xx

yym


Finding Slopes with the Slope Formula

Find the slope of the line through the given points.

(a) (–2, 4), (2, –6)

(b) (–3, 8), (5, 8)

(c) (–4, –10), (–4, 10)

the slope is undefined

Forms for the Equations of a Line

1.

uses slope, m and given point (x1, y1)

2. Slope-intercept form: y = mx + buses slope, m and y-intercept = b

3. Standard form: Ax +By = C

Slope is m = -A/B and (0, C/B) and (C/A,0)

4. Horizontal form: y = b

Slope is zero, y-interecept (0,b)

5. Vertical form: x = a

slope is undefined, not a function.


Point-slope form: y – y1 = m(x – x1)


Example 6 Finding the Slope from an Equation (page 222)

Find the slope of the line y = 5x – 3.

This equation is in slope-intercept form. You can identify the slope, m from its form.

m = 5


Graphing a Line Using a Point and the Slope

Graph the line passing through (–2, –3) and having slope .

Plot the point (–2, –3).

The slope is , so a change of 3 units horizontallyproduces a change of 4 units vertically. This gives the point (1, 1).

Join the points with a straight line to complete the graph.

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2-8

Graphing a Linear Function Using Intercepts

Graph . Give the domain and range.

Since this equation is in slope-intercept form, pull the slope from the equation, m = 3/2 and the y-intercept, (0,6).

Use the slope to move from

the point (0,6) up 3 units and

to the right 2 units.(0,6)

(2,9)

Domain:

Range:


Graphing a Linear Function Using Intercepts

Corresponding calculator graph:


Example 2 Graphing a Horizontal Line Graph f(x) = 2. Give the domain and range.

Since f(x) always equals 2, the value of y can never be 0. So, there is no x-intercept. The line is parallel to the x-axis.

The y-intercept is 2.

Domain: Range: {2}


Example 3 Graphing a Vertical Line Graph x = 5. Give the domain and range.

Since x always equals 5, the value of x can never be 0. So, there is no y-intercept. The line is parallel to the y-axis. The relation is not a function.

The x-intercept is 5.

Domain: {5} Range:


Graphing Ax + By = C with C = 0

Graph 3x + 4y = 0. Give the domain and range.

Find the intercepts.

x-intercept (x,0) 3x + 4(0) = 0

x = 0

y-intercept (0,y)3(0) + 4y = 0

y = 0

There is only one intercept, the origin, (0, 0). Choose x = 4 to find a second point.

3(4) + 4y = 012 + 4y = 0

y = –3


Graphing Ax + By = C with C = 0 (cont.)

Graph the points (0, 0) and (4, –3) and join with a straight line.

Domain: Range:

Average Rate of Change


We know that the slope of a line is the ratio of vertical change in y to the horizontal change in x. Thus, the slope gives the average rate of change in y per unit of change in x, where y = f(x). If f is a linear function we get the slope formula.

Average rate of change of f on a given interval, [a,b] is

ab

afbf

)()(

Types of Slopes

The slope indicates what is happening to a line on the graph:

a)If m > 0, the line will rise from left to right, i.e. it is increasing,

b)If m < 0, the line will fall from left to right, i.e. it is decreasing,

c)If m = 0, then there is no change in the line and it is horizontal, i.e. it is constant

d)If m is undefined, the line is a vertical line and not a function.



Example 8 Interpreting Slope as Average Rate of Change

In 1997, sales of VCRs numbered 16.7 million. In 2002, estimated sales of VCRs were 13.3 million. Find the average rate of change in VCR sales, in millions, per year. Graph as a line segment, and interpret the result.

The average rate of change per year is


Interpreting Slope as Average Rate of Change

In 1997, sales of VCRs numbered 16.7 million. In 2002, estimated sales of VCRs were 13.3 million. Find the average rate of change in VCR sales, in millions, per year. Graph as a line segment, and interpret the result.

The graph confirms that the line through the ordered pairs fall from left to right, and therefore has negative slope.

Sales of VCRs decreased by an average of 0.68 million each year from 1997 to 2002.


Writing Linear Cost, Revenue, and Profit Functions

Assume that the cost to produce an item is a linear function and all items produced are sold. The fixed cost is $2400, the variable cost per item is $120, and the item sells for $150. Write linear functions to model (a) cost, (b) revenue, and (c) profit.

(a) Since the cost function is linear, it will have the form C(x) = mx + b with m = 120 and b = 2400.

C(x) = 120x + 2400

(b) The revenue function is R(x) = px with p = 150.

R(x) = 150x



(c) The profit is the difference between the revenue and the cost.

= 150x – (120x + 2400)P(x) = R(x) – C(x)

= 30x – 2400

(d) How many items must be sold for the company to make a profit?

To make a profit, P(x) must be positive.

30x – 2400 > 0 30x > 2400 x > 80

The company must sell at least 81 items to make a profit.



Graphing calculator solution

Define Y1 as 30x – 2400 and graph the line. Then find the x-intercept.

The graph shows that y-values for x less than 80 are negative, y-values for x greater than 80 are positive. So, at least 81 items must be sold for the company to make a profit.

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Equations of LinesPoint-Slope Form ▪ Slope-Intercept Form ▪ Vertical and Horizontal Lines ▪ Parallel and Perpendicular Lines ▪ Modeling Data ▪ Modeling Data ▪ Solving Linear Equations in One Variable by Graphing

Forms for the Equations of a Line

1.

uses slope, m and given point (x1, y1)

2. Slope-intercept form: y = mx + buses slope, m and y-intercept = b

3. Standard form: Ax +By = C

Slope is m = -A/B and (0, C/B) and (C/A,0)

4. Horizontal form: y = b

Slope is zero, y-interecept (0,b)

5. Vertical form: x = a

slope is undefined, not a function.




Find an equation of the line through (3, –5) having slope –2.

Using the Point-Slope Form (Given a Point and the Slope) (page 232)


x1 = 3, y1 = –5, m = –2


Find an equation of the line through (–4, 3) and (5, –1).

2.5 Example 2 Using the Point-Slope Form (Given Two Points) (page 233)

First, find the slope:

Use either point for (x1, y1)

Point-slope form


Find an equation of the line through (–4, 3) and (5, –1).

Using the Point-Slope Form (Given Two Points)

Verify using (5, –1) for (x1, y1) :

Point-slope form


Find the slope and y-intercept of the line with equation 3x – 4y = 12.

Find the Slope and y-intercept From an Equation of a Line

Write the equation in slope-intercept form:

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


Find an equation of the line through (–2, 4) and (2, 2). Then graph the line using the slope-intercept form.

Using the Slope-Intercept Form (Given Two Points)

First, find the slope:

The equation is

Substitute for m and the coordinates of one of the

points (say, (2, 2)) for x and y into the slope-intercept

form y = mx + b, then solve for b:


Using the Slope-Intercept Form (Given Two Points)


Use the graph to (a) find the slope, y-intercept, and x-intercept, and (b) write the equation of the function.

Finding an Equation From a Graph

The line rises 5 units each time the x-value increases by 2 units.

The slope is .


2.5 Example 5(a) Finding an Equation From a Graph (cont.)

The graph intersects the y-axis at (0, 5) and the x-axis at (–2, 0).

The y-intercept is 5.

The x-intercept is –2.


Finding an Equation From a Graph

Slope , y-intercept 5

Parallel and Perpendicular Lines

Parallel Lines:

Two distinct non-vertical lines are parallel if and only if they have the same slopes. In other words, if line l1 has slope of m1 and l2 has slope of m2 and l1 is parallel to l2, then

Perpendicular Lines:

The product of the slopes of perpendicular lines is -1. In other words, if line l1 has slope of m1 and l2 has slope of m2 and l1 is perpendicular to l2, then


21 mm

12

1

mm


Find the equation in slope-intercept form of the line that passes through the point (2, –4) that is parallel to the line 3x – 2y = 5.

Finding Equations of Parallel and Perpendicular Lines

Write the equation in slope-intercept form to find the slope:

The slope is .

Parallel lines have the same slope, so the slope of the line whose equation is to be found is .


Find the equation in slope-intercept form of the line that passes through the point (2, –4) that is parallel to the line 3x – 2y = 5.


Point-slope form


Find the equation in slope-intercept form of the line that passes through the point (2, –4) that is perpendicular to the line 3x – 2y = 5.


Write the equation in slope-intercept form to find the slope:

The slope is .

The slopes of perpendicular lines are negative reciprocals, so the slope of the line whose equation is to be found is .


Find the equation in slope-intercept form of the line that passes through the point (2, –4) that is perpendicular to the line 3x – 2y = 5.


Point-slope form


Average annual tuition and fees for in-state students at public 4-year colleges are

shown in the table for selected years and in the graph below, with x = 0 representing 1996, x = 4 representing 2000, etc.

Finding an Equation of a Line That Models Data


Find an equation that models the data. Use the data for 1998 and 2004.


1998 is represented by x = 2 and 2004 is represented by x = 8.

Find the slope:



Use either point, (2, 3486) or (8, 5148) for (x1, y1)



Use the equation from part (a) to predict the cost of tuition and fees in 2008.

For 2008, x = 12.

According to the model, average tuition and fees will be $6256 in 2008.


The table and graph illustrate how the percent of women in the civilian labor force has changed from 1960 to 2005.



Use the points (1965, 39.3) and (1995, 58.9) to find a linear equation that models the data.


Find the slope:

Use either point for (x1, y1).


Use the equation to estimate the percent for 2005. How does the result compare to the actual figure of 59.3%?


Let x = 2005. Solve for y:

The model estimates about 65.4% in 2005.

This is 6.1% more than the actual figure of 59.3%.

Linear Equations Graphs and Functions. Copyright © 2008 Pearson Addison-Wesley. All rights...

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