1 Copyright © 2011 Pearson Education, Inc.. Equations and Inequalities in Two Variables; Functions...

1

Copyright © 2011 Pearson Education, Inc.


Equations and Inequalities in Two Variables; Functions

CHAPTER

3.1 Graphing Linear Equations3.2 The Slope of a Line3.3 The Equation of a Line3.4 Graphing Linear Inequalities3.5 Introduction to Functions and

Function Notation

33


The Slope of a Line3.23.2

1. Compare lines with different slopes.2. Graph equations in slope-intercept form.3. Find the slope of a line given two points on the line.

4Copyright © 2011 Pearson Education, Inc.

Example 1a

Graph each of the following on the same grid.y = x y = 3x y = 4xSolutionComplete a table of values.

If x is y = x y = 3x y = 4x

0 0 0 0

1 1 3 4

2 2 6 8 y = x

y = 3xy = x

y = 3xy = x

y = 4x


Example 1b

Graph each of the following on the same grid.

SolutionComplete a table of values.

1 1

3 4y x y x y x

If x is y = x

0 0 0 0

1 1

2 2

13-y x 1

4-y x

13

14

23

12


If the coefficient of m increases, the graphs get steeper. Because the coefficient m affects how steep a line is, m is called the slope of the line.

If a slope of the line is a fraction between 0 and 1, then the smaller the fraction is, the less inclined or flatter the line gets.


Slope: The ratio of the vertical change (change in y) to the horizontal change (change in x) between any two points on a line between those points.


Example 2

For the equation determine the slope and the y-intercept. Then graph the equation.Solutionm =

y-intercept: (0, 3)

Plot the y-intercept and then use the slope to find other points.

23,

3y x

2

3

rise 2

run 3(3, 1)


Graphs of Equations in Slope-Intercept Form The graph of an equation in the form y = mx + b, (slope-intercept form) is a line with slope m and y-intercept (0, b). The following rules indicate how m affects the graph.

If m > 0, the line slants upward from left to right.If m < 0, the line slants downward from left to right.The greater the absolute value of m, the steeper the line.

m > 0

m < 0


Example 3

For the equation 2x + 5y = 20, determine the slope and the y-intercept. Then graph the equation.SolutionWrite the equation in slope-intercept form by isolating y.2x + 5y = 205y = 2x 20

2 20

5 5y x

24

5y x

The slope is and the y-intercept is (0, –4).

25


continued

We begin at (0, –4) and then rise 2 and run 5.

m = y-intercept: (0, –4)

25

rise 2 run 5


The Slope Formula

Given two points (x1, y1) and (x2, y2), where x2 x1, the slope of the line connecting the two points is given by the formula

2 1

2 1

.y y

mx x

Rise:

y2 – y1

Run:

x2 – x1(x1, y1)

(x2, y2)


Find the slope of the line connecting (4, 6) and (−2, 8).

Solution

Example 4

Using , replace the variables with their corresponding values and then simplify.

2 1

2 1

y y

mx x

2 1

2 1

y y

mx x

8 6

2 4m

2

6m

1

3


Graph the line connecting the given points and find its slope. (3, 8) and (−2, 8)

Solution

Example 5a

Because the y-coordinates are the same, the graphs is a horizontal line.

8 8

2 3m

0

5m

0


Graph the line connecting the given points and find its slope. (6, 1) and (6, −4)

Solution

Example 5b

Because the x-coordinates are the same, the graphs is a vertical line.

4 1

6 6m

5

, which is undefined.0

m

Graph the line connecting the given points and find its slope. (6, 1) and (6, −4)

X

Y

-10 -8 -6 -4 -2 2 4 6 8 10

-10-8

-6-4

-2

2

468

10

0


Slopes of Horizontal and Vertical Lines

Two points with different x-coordinates and the same y-coordinates, (x1, c) and (x2, c), will form a horizontal line with slope 0 and equation y = c.

Two points with the same x-coordinates and different y-coordinates, (c, y1) and (c, y2), will form a vertical line with undefined slope and equation x = c.


Example 6The following graph shows the hourly wage earned by Henry each of the five years after his hire date. An analysis determines that the red line reasonably describes the trend shown in the data. Find the slope of that line.

Understand We are to find the slope on a line that passes through or near a set of data points.

X

Y

1 2 3 4 5

2

4

6

8

10

12

14

16

18

20

0


continued

Execute18 15

5 3m

3

2

Plan To find the slope of the line, we use the slopeformula with two data points that are on the line.

We will use (3, 15) and (5, 18).

2 1

2 1

y y

mx x

Answer The slope of the line is , which means that Henry’s wage increased about $1.50 per hour each year.

3

2

Check We could use a different pair of data points on the line and see if we get the same slope. We will leave this to the viewer.


What is the slope of the line 3x + y = 6?

a) m = 3

b) m = 3

c)

d) m = 2

1

2m

3.2


How would you graph the line y = 3x – 2?

a) Plot (0, –2), down 2, right 3

b) Plot (0, –2), down 3, right 1

c) Plot (0, 2), up 3, left 2

d) Plot (0, –2), up 3, right 1

3.2


What is the slope of the line through the given points (5, 6) and (6, 3)?

a) m = 9

b) m = 9

c)

d)

1

9m

9

11m

3.2


The Equation of a Line3.33.3

1. Use slope-intercept form to write the equation of a line.2. Use point-slope form to write the equation of a line.3. Write the equation of a line parallel to a given line.4. Write the equation of a line perpendicular to a given line.


If we are given the y-intercept (0, b) of a line, to write the equation of the line, we will need either the slope or another point so that we can calculate the slope.


Example 1

A line with a slope of 4 crosses the y-axis at the point (0, 5). Write the equation of the line in slope-intercept form.SolutionUse y = mx + bm = 4b = 5 from the point given (0, 5) which is the y-intercept.

y = 4x + 5


Equation of a Line Given Its y-intercept

To write the equation of a line given its y-intercept, (0, b), and its slope, m, use the slope-intercept form of the equation, y = mx + b. If given a second point and not the slope, calculate the slope using then use y = mx + b.

2 1

2 1

;y y

mx x


Example 2Write the equation of the line passing through (0, 6) and (3, 6) in slope-intercept form.SolutionThe y-intercept is (0, 6).Find the slope:

2 1

2 1

y ym

x x

6 ( 6)

3 0m

12

3m 4

y = mx + by = 4x 6


You can use the point-slope form to write the equation of a line given any two points on the line.

Using the Point-Slope Form of the Equation of a LineTo write the equation of a line given its slope and any point, (x1, y1), on the line, use the point-slope form of the equation of a line, y – y1 = m(x – x1). If given a second point (x2, y2), and not the slope, we first calculate the slope using then use y – y1= m(x – x1).

2 1

2 1

,y y

mx x


Example 3aWrite the equation of a line with a slope of 5 that passes through the point (3, 12). Write the equation in slope-intercept form.SolutionBegin with the point-slope formula.Replace m = 5, x1 = 3, y1 = 12

y – y1 = m(x – x1)

y – 12 = 5(x – 3)y – 12 = 5x – 15 Simplify.

y = 5x – 3 Add 15 to both sides to isolate y.


Example 3bWrite the equation of a line passing through the points (4, 2) and (4, 4). Write the equation in slope-intercept form.SolutionCalculate the slope.

Use the point-slope form, then isolate y to write the slope-intercept form.

2 1

2 1

4 ( 2)

4 4

y ym

x x

6 3

8 4


continuedPoints (4, 2) and (4, 4),

y – y1 = m(x – x1)

3

4m

3( 2) ( 4)

4y x

32 3

4y x

31

4y x

Replace m with -3/4, x1 with 4 and y1 with -2.

Simplify.

Subtract 2 from both sides to isolate y.


Equations can also be written in standard form, which is Ax + By = C, where A, B, and C are real numbers.

It should also be written so that the x-term is first and with a positive coefficient and, if possible, A, B, and C are all integers.


Example 4A line connects the points (2, 6) and (–4, 3). Write the equation of the line in the form Ax + By = C, where A, B, and C are integers and A > 0.SolutionFind the slope:

Use point-slope form: y – y1 = m(x – x1)

2 1

2 1

3 6

4 2

y ym

x x

3

6

1

2

16 ( 2)

2y x

16 1

2y x Distribute to clear ( ).


continued

16 1

2y x

12 6 2 1

2y x

2 12 2y x

2 12 2x y

2 10x y

Multiply both sides by the LCD, 2.

Subtract x from both sides to get x and y together.

Add 12 to both sides to get the constant terms together.

Multiply by –1 so that the coefficient of x is positive.

( 1)( 2 ) ( 1)(10)x y

2 10x y


Parallel LinesNonvertical parallel lines have equal slopes and different y-intercepts. Vertical lines are parallel.

y = 2x + 1 y = 2x – 3


Example 5Write the equation of the line in slope-intercept form that passes through (1, –5) and parallel to the graph of y = –3x + 4. SolutionIn y = –3x + 4, the slope is –3, so the slope of the line parallel will also be –3.Use point-slope form. y – y1 = m(x – x1)

y – (5) = –3(x – 1) y + 5 = –3x + 3

y = –3x – 2

y1 = 5, x1 = 1 and m = –3

Simplify.

Subtract 5 from both sides to isolate y.


Perpendicular LinesThe slope of a line perpendicular to a line with a slope of will be Horizontal and vertical lines are perpendicular.

a

b.

b

a

24

3y x

31

2y x


Example 6Write the equation of a line in standard form that passes through (7, 1) and is perpendicular to the graph of 7x – 2y = –2. SolutionDetermine the slope of the line 7x – 2y = –2.

Slope of perpendicular line:

2 7 2y x 7

12

y x 2

7


continued slope = ; point (7, 1)

y – y1 = m(x – x1)

Multiply both sides by 7, simplify and distribute.

Rearrange the x and y terms, and add 7 to both sides to put the equation in Standard Form.

21 ( 7)

7y x

7 7 2 14y x

2

7

2 7 7 14x y

2 7 21x y


Write the equation of the line in slope-intercept form given m = 2 and the point (4, 5).

a) y = 2x – 3

b) y = 2x + 3

c) y = 2x + 3

d) y = 2x – 5

3.3


What is the equation of the line connecting the points (4, 3) and (1, 7)?

a) y = 2x + 5

b) y = 2x + 5

c) y = 2x – 5

d) 1

52

y x

3.3


What is the relationship between the two lines? 5x – 3y = 11

3x + 5y = 8

a) parallel

b) perpendicular

c) neither

3.3


Introduction to Functions and Function Notation3.53.5

1. Identify the domain and range of a relation and determine whether a relation is a function.

2. Find the value of a function.3. Graph functions.


Relation: A set of ordered pairs.

Domain: The set containing initial values of a relation; its input values; the first coordinates in ordered pairs.

Range: The set containing all values that are paired to domain values in a relation; its output values; the second coordinates in ordered pairs.


Function: A relation in which each value in the domain is assigned to exactly one value in the range.

Domain Range 0 2 1 4 2 6 3 8 4 10

Each element in the domain has a single arrow pointing to an element in the range.


Every function is a relation, but not every relation is a function.

If any value in the domain is assigned to more than one value in the range, then the relation is not a function.Domain Range 0 2 1 4 2 6

1012

not a function


Example 1Identify the domain and range of the relation, then determine if it is a function. Birthdate Family memberMarch 1 DonnaApril 17 DennisSept. 3 CatherineOctober 9 Denise

Nancy

The relation is not a function because an element in the domain, Sept. 3, is assigned to two names in the range.

Domain: {March 1, April 17, Sept 3, Oct 9}

Range: {Donna, Dennis, Catherine, Denise, Nancy}


Determining the Domain and Range of a GraphThe domain is a set containing the first coordinate (x-coordinate) of every point on the graph.

The range is a set containing the second coordinate (y-coordinate) of every point on the graph.


Vertical Line TestTo determine whether a graphical relation is a function, draw or imagine vertical lines through each value in the domain. If each vertical line intersects the graph at only one point, the relation is a function. If any vertical line intersects the graph more than once, the relation is not a function.


Example 2

For each graph, identify the domain and range. Then state whether each relation is a function.a. b.

Domain: {x|x 1}Range: all real numbers

Not a functionDomain: all real numbersRange: {y 1}

Function


When written as an equation, the notation for a function is a modification of an equation in two variables.

y = 3x + 4 could be written as f(x) = 3x + 4

f(x) is read as “a function in terms of x” or “f of x”


Finding the Value of a Function

Given a function f(x), to find f(a), where a is a real number in the domain of f, replace x in the function with a and then evaluate or simplify.


Example 3

For the function f(x) = 3x – 5, find the following.a. f(2) b. f(4) c. f(a)

Solutiona. f(2) = 3x – 5

= 3(2) – 5 = 6 – 5 = 1

b. f(4) = 3x – 5

= 3(4) – 5

= 12 – 5

= 17

c. f(a) = 3x – 5

= 3(a) – 5

= 3a – 5


Example 4

Use the graph to find the indicated value of the function.a. f(0) b. f(2) c. f(8)

Solutiona. When x = 0, y = 0, so f(0) = 0.

b. When x = 2, y = 2, so f(2) = 2.

X

Y

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

0

c. When x = 8, y = 4, so f(8) = 4.


We create the graph of a function the same way that we create the graph of an equation in two variables.

Slope-intercept form: y = mx + bLinear function: f(x) = mx + b


Example 5

Graph: f(x) = 2x + 1. Give the domain and range.SolutionWe could make a table of values or use the fact that the slope is 2 and the y-intercept is 1.

Domain: Range:

x f(x)

0 1

1 3

2 5

f(x) = 2x + 1

or ( , )

or ( , )


Example 7Graph. f(x) = 2x2

Solution We create a table of ordered pairs, plotthe points, and connect with a smooth curve.

x f(x)

2 8

1 2

0 0

1 2

2 8


Is the relation a function?

a) yes

b) no

3.5


For the function f(x) = 2x2 + x – 4 find f(1).

a) 3

b) –4

c) 2

d) 1

3.5

1 Copyright © 2011 Pearson Education, Inc.. Equations and Inequalities in Two Variables; Functions...

Documents

Transcript of 1 Copyright © 2011 Pearson Education, Inc.. Equations and Inequalities in Two Variables; Functions...