The Coordinate Plane

Pre-AlgebraPre-Algebra


The Coordinate PlaneThe Coordinate Plane

Lesson 1-10

Write the coordinates of point G. In which

quadrant is point G located?

Point G is located 2 units to the left of the y-axis.So the x-coordinate is –2.

The point is 3 units below the x-axis.

So the y-coordinate is –3.

The coordinates of point G are (–2, –3). Point G is located in Quadrant III.


The Coordinate PlaneThe Coordinate Plane

Lesson 1-10

Graph point M(–3, 3).


Relations and FunctionsRelations and Functions

Lesson 8-1

Is each relation a function? Explain.

a. {(0, 5), (1, 6), (2, 4), (3, 7)}

List the domain values and the range values in order.

Draw arrows from the domain values to their range values.

There is one range value for each domain value. This relation is a function.



Lesson 8-1

(continued)

b. {(0, 5), (1, 5), (2, 6), (3, 7)}

There is one range value for each domain value. This relation is a function.



Lesson 8-1

(continued)

c. {(0, 5), (0, 6), (1, 6), (2, 7)}

There are two range values for the domain value 0.

This relation is not a function.



Lesson 8-1

Is the time needed to mow a lawn a function of the

size of the lawn? Explain.

No; two lawns of the same size (domain value) can require different lengths of time (range values) for mowing.



Lesson 8-1

a. Graph the relation shown in the table.

Domain Value

–3

–5

3

5

Range Value

5

3

5

3

Graph the ordered pairs(–3, 5), (–5, 3), (3, 5), and (5, 3).



Lesson 8-1

b. Use the vertical-line test. Is the relation a function? Explain.

(continued)

The pencil does not pass through two points at any one of its positions, so the relation is a function.

Pass a pencil across the graph as shown. Keep the pencil vertical (parallel to the y-axis) to represent a vertical line.

Pre-AlgebraPre-AlgebraLesson 8-2

Find the solution of y = 4x – 3 for x = 2.

y = 4x – 3

y = 4(2) – 3 Replace x with 2.

y = 8 – 3 Multiply.

y = 5 Subtract.

A solution of the equation is (2, 5).

Equations With Two VariablesEquations With Two Variables


The equation a = 5 + 3p gives the price for

admission to a park. In the equation, a is the admission

price for one car with p people in it. Find the price of

admission for a car with 4 people in it.

a = 5 + 3p

a = 5 + 3(4) Replace p with 4.

a = 5 + 12 Multiply.

a = 17 Add.

A solution of the equation is (4, 17). The admission price for one car with 4 people in it is $17.



x 4x – 2 (x, y)

–2 4(–2) – 2 = –8 – 2 = –10 (–2, –10)

0 4(0) – 2 = 0 – 2 = –2 (0, –2)

2 4(2) – 2 = 8 – 2 = 6 (2, 6)

Graph the ordered pairs.

Draw a line through the points.

Lesson 8-2

Graph y = 4x – 2.

Make a table of values to show ordered-pair solutions.



For every value of x, y = –3.

Lesson 8-2

Graph each equation. Is the equation a function?

This is a horizontal line.

a. y = –3 b. x = 4

The equation y = –3 is a function.

This is a vertical line.The equation y = 4 is not a function.

For every value of y, x = 4.



Solve y – x = 3 for y. Then graph the equation.12

Solve the equation for y.

y – x = 312

y = x + 3 Simplify.12

y – x + x = 3 + x Add x to each side.12

12

12

12



(continued)

Graph.Make a table of values.

x x + 3 (x, y)

–2 (–2) + 3 = –1 + 3 = 2 (–2, 2)

0 (0) + 3 = 0 + 3 = 3 (0, 3)

2 (2) + 3 = 1 + 3 = 4 (2, 4)

12121212



Slope and y-interceptSlope and y-intercept

Lesson 8-3



Lesson 8-3

Find the slope of each line.

a. b.

slope = = = 4riserun

41 slope = = = –2

riserun

–63



Lesson 8-3

Find the slope of the line through E(7, 5) and F(–2, 0).

slope = = = =difference in y-coordinatesdifference in x-coordinates

0 – 5–2 – 7

–5–9

59



Lesson 8-3

Find the slope of each line.

a. b.

slope = –3 – (–3)4 – (–2) = = 0

06

Slope is 0 for a horizontal line.

slope = –1 – 3

–2 – (–2) =–40

Division by zero is undefined. Slope is undefined for a vertical line.



Lesson 8-3

A ramp slopes from a warehouse door down to a

street. The function y = – x + 4 models the ramp, where x is

the distance in feet from the bottom of the door and y is the

height in feet above the street. Graph the equation.

15

Step 1: Since the y-intercept is 4, graph (0, 4).

Step 3: Draw a line through the points.

Then move 5 units right to graph a second point.

Step 2: Since the slope is – , move

1 unit down from (0, 4).

15


Writing Rules for Linear FunctionsWriting Rules for Linear Functions

Lesson 8-4

A long-distance phone company charges its

customers a monthly fee of $4.95 plus 9¢ for each minute of

a long-distance call.

a. Write a function rule that relates the total monthly bill to the number of minutes a customer spent on long-distance calls.

A rule for the function is t(m) = 4.95 + 0.09m.

Words total bill is $4.95 plus 9¢ timesnumber of minutes

Let = the number of minutes.m

Let = total bill, a function of the number of minutes.t( m )

Rule t( m ) = 4.95 + • m0.09



Lesson 8-4

(continued)

b. Find the total monthly bill if the customer made 90 minutes of long-distance calls.

t(m) = 4.95 + 0.09m

The total monthly bill with 90 minutes of long-distance calls is $13.05.

t(90) = 4.95 + 0.09(90) Replace m with 90.

t(90) = 4.95 + 8.10 Multiply.

t(90) = 13.05 Add.



Lesson 8-4

Write a rule for the linear function in the table below.

x

2

0

–2

–4

f(x)

3

–5

–13

–21

–8

–8

–8

–2

–2

–2

As the x values decrease by 2, the f(x) values decrease by 8.

So m = = 4.–8–2

When x = 0, f(x) = –5. So b = –5.

A rule for the function is f(x) = 4x – 5.



Lesson 8-4

Write a rule for the linear function graphed below.

slope = –2 – 20 – 2 =

–4–2 = 2

y-intercept = –2

A rule for the function is f(x) = 2x – 2.


Scatter PlotsScatter Plots

Lesson 8-5

The scatter plot shows education and income data.

a. Describe the person represented by the point with coordinates (10, 30).This person has 10 years of education and earns $30,000 each year.

b. How many people have exactly 14 years of education? What are their incomes?The points (14, 50), (14, 80), and (14, 90) have education coordinate 14.The three people they represent earn $50,000, $80,000, and $90,000, respectively.



Lesson 8-5

Use the table to make a scatter plot of the

elevation and precipitation data.

Atlanta, GA

Boston, MA

Chicago, IL

Honolulu, HI

Miami, FL

Phoenix, AZ

Portland, ME

San Diego, CA

Wichita, KS

Elevation Above Sea Level (ft)

1,050

20

596

18

11

1,072

75

40

1,305

CityMean Annual

Precipitation (in.)

51

42

36

22

56

8

44

10

29



Lesson 8-5

Use the scatter plot below. Is there a positive

correlation, a negative correlation, or no correlation between

temperatures and amounts of precipitation? Explain.

The values show no relationship.

There is no correlation.


Problem Solving Strategy: Solve by Graphing Problem Solving Strategy: Solve by Graphing

Lesson 8-6

Use the data in the table below. Suppose this year

there are 16 wolves on the island. Predict how many moose

are on the island.

Isle Royale Populations

Wolf Moose

14

23

24

22

20

16

700

900

811

1,062

1,025

1,380

Year

1982

1983

1984

1985

1986

1987

Wolf Moose

12

11

15

12

12

13

1,653

1,397

1,216

1,313

1,600

1,880

Year

1988

1989

1990

1991

1992

1993

Wolf Moose

15

16

22

24

14

25

1,800

2,400

1,200

500

700

750

Year

1994

1995

1996

1997

1998

1999


Step 1: Make a scatter plot by graphing the (wolf, moose) ordered pairs. Use the x-axis for wolves and the y-axis for moose.

Step 2: Sketch a trend line. The line should be as close as possible to each data point. There should be about as many points above the trend line as below it.


Lesson 8-6

(continued)


Look up to find the point on the trend line that corresponds to 16 wolves.

Then look across to the value on the vertical axis, which is about 1,300.


Lesson 8-6

(continued)

Step 3: To predict the number of moose when there are 16 wolves, find 16 along the horizontal axis.

There are about 1,300 moose on the island.


Solve the system y = x – 7 and y = 4x + 2 by

graphing.

Solving Systems of Linear EquationsSolving Systems of Linear Equations

Lesson 8-7

Step 1: Graph each line.Step 2: Find the point of intersection.

The lines intersect at one point, (–3, –10). The solution is (–3, –10).

y = x – 7

Check: See whether (–3, –10) makes both equations true.

–10 –3 – 7

–10 = –10

y = 4x + 2

–10 4(–3) + 2

–10 = –10

Replace x with – 3and y with –10.

The solution checks.


Solve each system of equations by graphing.


Lesson 8-7

a. 27x + 9y = 36; y = 4 – 3x b. 8 = 4x + 2y; 2x + y = 5

The lines are the same line.There are infinitely many solutions.

The lines are parallel.They do not intersect.There is no solution.


Find two numbers with a sum of 10 and a

difference of 2.


Lesson 8-7

Step 1: Write equations.Let x = the greater number.Let y = the lesser number.

Equation 1 Sum is 10.x + y = 10

Equation 2 Difference is 2.x – y = 2

Step 2: Graph the equations.The lines intersect at (6, 4).The numbers are 6 and 4.


(continued)


Lesson 8-7

Check: Since the sum of 6 and 4 is 10 and the difference of 6 and 4 is 2, the answer is correct.


Graph each inequality on a coordinate plane.

Graphing Linear InequalitiesGraphing Linear Inequalities

Lesson 8-8

a. y > 2x + 1

Step 1: Graph the boundary line.Points on the boundary line do not make y > 2x + 1 true. Use a dashed line.


(continued)


Step 2: Test a point not on the boundary line.Test (0, 0) in the inequality.

Since the inequality is false for (0, 0), shade the region that does not contain (0, 0).

y > 2x + 10 2(0) + 1 Substitute.0 0 + 1

0 > 1 false

>?

>?

Lesson 8-8


(continued)


b. y 3x – 2<–Step 1: Graph the boundary line.

Points on the boundary line make y 3x – 2 true. Use a solid line.<–

Lesson 8-8


(continued)


Step 2: Test a point not on the boundary line.Test (3, 0) in the inequality.

Since the inequality is true for (3, 0), shade the region containing (3, 0).

y 3x – 20 3(3) – 2 Substitute.0 9 – 20 7 true

<–

<–

<–?

<–?

Lesson 8-8


Cashews cost $2/lb. Pecans cost $4/lb. You plan

to spend no more than $20. How many pounds of each can

you buy?


Step 1: Write an inequality.

Wordscost of

cashewsplus

cost of pecans

is at most

twenty dollars

Let = number of pounds of pecans.x

+Inequality 2y 4x <– 20

Let = number of pounds of cashews.y

Lesson 8-8


(continued)


Step 3: Graph y = –2x + 10 in Quadrant I since weight is not negative.

Lesson 8-8

Step 2: Write the equation of the boundary line in slope-intercept form.

2y + 4x 20

y –2x + 10

<–

y = –2x + 10

<–


(continued)


Step 4: Test (1, 1).y –2x + 101 –2(1) + 101 8The inequality is true. (1, 1) is a solution.

<–<–?

<–

Step 5: Shade the region containing (1, 1).

The graph shows the possible solutions. For example, you could buy 1 pound of pecans and 5 pounds of cashews.

Lesson 8-8


Step 1: Graph y x + 1 on a coordinate plane. Shade in red.

>–

Solve the system y x + 1 and y < 2x + 3 by graphing.


>–

Step 2: Graph y < 2x + 3 on the same coordinate plane. Shade in blue.

Lesson 8-8


(continued)


The solutions are the coordinates of all the points in the region that is shaded in both colors.

Check: See whether the solution (2, 5) makes both of the inequalities true.y x + 1 5 2 + 1 Replace x with 2 and y with 5.5 3y The solution checks.

y < 2x + 3 5 2(2) + 3 Replace x with 2 and y with 5.5 < 7y The solution checks.

>–>–?

>–

<?

Lesson 8-8

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The Coordinate Plane