Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

OBJECTIVES

Graphs of Equations

Sketch a graph by plotting points.

Find the intercepts of a graph.

Find the symmetries in a graph.

Find the equation of a circle.

SECTION 2.2

1

2

3

4

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DefinitionsAn ordered pair (a, b) is said to satisfy an equation with variables a and b if, when a is substituted for x and b is substituted for y in the equation, the resulting statement is true.

An ordered pair that satisfies an equation is called a solution of the equation.

Frequently, the numerical values of the variable y can be determined by assigning appropriate values to the variable x. For this reason, y is sometimes referred to as the dependent variable and x as the independent variable.

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GRAPH OF AN EQUATION

The graph of an equation in two variables, such as x and y, is the set of all ordered pairs (a, b) in the coordinate plane that satisfy the equation.

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EXAMPLE 1 Sketching a Graph by Plotting Points

Sketch the graph of y x2 3.

Solution

(3, 6)y = 32 – 3 = 9 – 3 = 63

(2, 1)y = 22 – 3 = 4 – 3 = 12

(1, –2)y = 12 – 3 = 1 – 3 = –21

(0, –3)y = 02 – 3 = 0 – 3 = –30

(–1, –2)y = (–1)2 – 3 = 1 – 3 = –2–1

(–2, –1)y = (–2)2 – 3 = 4 – 3 = 1–2

(–3, 6)y = (–3)2 – 3 = 9 – 3 = 6–3

(x, y)y = x2 – 3x

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EXAMPLE 1 Sketching a Graph by Plotting Points

Solutioncontinued

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SKETCHING A GRAPH BY PLOTTING POINTS

Step1. Make a representative table of solutions of the equation.

Step 2. Plot the solutions as ordered pairs in the Cartesian coordinate plane.

Step 3. Connect the solutions in Step 2 by a smooth curve.

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Definitions

The points where a graph intersects (crosses or touches) the coordinate axes are of special interest in many problems. Since all points on the x-axis have a y-coordinate of 0, any point where a graph intersects the x-axis has the form (a, 0). The number a is called an x-intercept of the graph. Similarly, any point where a graph intersects the y-axis has the form (0, b), and the number b is called a y-intercept of the graph.

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PROCEDURE FOR FINDING THE INTERCEPTS OF A GRAPH

Step1 To find the x-intercepts of an equation, set y = 0 in the equation and solve for x.

Step 2 To find the y-intercepts of an equation, set x = 0 in the equation and solve for y.

Slide 2.2- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 2 Finding Intercepts

Find the x- and y-intercepts of the graph of the equation y = x2 – x – 2.

SolutionStep 1 To find the x-intercepts, set y = 0, solve for x.

The x-intercepts are –1 and 2.

0 x2 x 2

0 x 1 x 2 x 1 0 or x 2 0

x 1 or x 2

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EXAMPLE 2 Finding Intercepts

Solution continued

Step 2 To find the y-intercepts, set x = 0, solve for y.

y 02 0 2

y 2

The y-intercept is –2.

This is the graph of the equation y = x2 – x –2.

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Definitions

A useful tool for sketching the graph of an equation is the concept of symmetry, which means that one portion of the graph is a mirrorimage of another portion. The mirror line is usually called the axis of symmetry or line of symmetry.

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TESTS FOR SYMMETRY

1. A graph is symmetric with respect to the y-axis if, for everypoint (x, y) on the graph,the point (–x, y)is also on the graph.

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TESTS FOR SYMMETRY

2. A graph is symmetric with respect to the x-axis if, for everypoint (x, y) on the graph,the point (x, –y)is also on the graph.

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TESTS FOR SYMMETRY

3. A graph is symmetric with respect to the origin if, for everypoint (x, y) on the graph,the point (–x, –y)is also on the graph.

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EXAMPLE 3

Determine whether the graph of the equation

Checking for Symmetry

Solution

Replace x with –x in the original equation.

y 1

x2 5is symmetric with respect to the y-axis.

y 1

–x 2 5

y 1

x2 5

When we replace x with –x in the equation, we obtain the original equation. The graph is symmetric with respect to the y-axis.

Slide 2.2- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 4

Show that the graph of

Checking for Symmetry

Solution

Replace y with –y in the original equation.

x3 y2 xy2 0

is symmetric with respect to the x-axis.

When we replace y with –y in the equation, we obtain the original equation. The graph is symmetric with respect to the x-axis.

x3 y 2 x y 2 0

x3 y2 xy2 0

Slide 2.2- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 5

Show that the graph of

Checking for Symmetry with Respect to the Origin

Solution

Replace x with –x and y with –y .

y5 x3

with respect to the origin, but not with respect to either axis.

Replacing x with –x and y with –y in the equation, we obtain the original equation. The graph is symmetric with respect to the origin.

is symmetric

y 5 x 3

y5 x3

y5 x3

Slide 2.2- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 5Checking for Symmetry with Respect to the Origin

Solution continued

The graph of y5 = x3 is not symmetric with respect to the y-axis because the point (1, 1) is on the graph, but the point (–1, 1) is not on the graph. Similarly, the graph of y5 = x3 is not symmetric with respect to the x-axis since the point (1, 1) is on the graph, but the point (1, –1) is not on the graph.

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EXAMPLE 6Sketching a Graph by Using Intercepts and Symmetry

Initially, there are 400 deer on Dooms island. The number y of deer on the island after t years is described by the equation

a. Sketch the graph of the equation

y t 4 96t 2 400.

y t 4 96t 2 400.

b. Adjust the graph in part (a) to account for only the physical aspects of the problem.

c. When does the population of deer become extinct on Dooms?

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EXAMPLE 6Sketching a Graph by Using Intercepts and Symmetry

Solution continued

a. Find all intercepts.

y 04 96 0 2 400 400Set y = 0.

The t-intercepts are –10 and 10.

0 t 4 96t 2 400

t 4 96t 2 400 0

t 2 4 t 2 100 0

t 2 4 t 10 t 10 0

t 2 4 0 or t 10 0 or t 10 0

t 10, or t 10

y-intercept is 400.Set t = 0.

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EXAMPLE 6Sketching a Graph by Using Intercepts and Symmetry

Solution continued

Check for symmetry (t replaces x).

y t 4 96t 2 400

Symmetry in the t-axis: replace y with –y

(0, – 400) is a solution of

y t 4 96t 2 400.but not of

y t 4 96t 2 400

The graph is not symmetric in the t-axis.

Slide 2.2- 22 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 6Sketching a Graph by Using Intercepts and Symmetry

Solution continued

y t 4 96 t 2 400

y t 4 96t 2 400

Symmetry in the y-axis: replace t with –t

This is the original equation. Thus, (–t, y) also satisfies the equation and the graph is symmetric in the y-axis.

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EXAMPLE 6Sketching a Graph by Using Intercepts and Symmetry

Solution continued

y t 4 96 t 2 400

y t 4 96t 2 400

Symmetry in the origin: replace t with –t and y with –y

(0, 400) is a solution but (–0, –400) is not a solution, the graph is not symmetric with respect to the origin.

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EXAMPLE 6Sketching a Graph by Using Intercepts and Symmetry

Solution continued

Plot points for t ≥ 0 and then using symmetry in the y-axis.

t y = – t 4 +96t

2 +400 (t, y)

0 400 (0, 400)

1 495 (1, 495)

5 2175 (5, 2175)

7 2703 (7, 2703)

9 1615 (9, 1615)

10 0 (10, 0)

11 –2625 (11, –2625)

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EXAMPLE 6Sketching a Graph by Using Intercepts and Symmetry

Solution continued

Slide 2.2- 26 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 6Sketching a Graph by Using Intercepts and Symmetry

Solution continued

b. The graph pertaining to the physical aspects of the problem is the blue portion.

c. The positive t-intercept, which is 10, gives the time in years when the deer population of Dooms is 0.

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CIRCLE

A circle is a set of points in a Cartesian coordinate plane that are at a fixed distance r from a specified point (h, k). The fixed distance r is called the radius of the circle, and the specified point (h, k) is called the center of the circle.

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CIRCLEThe graph of a circle with center (h, k) and radius r.

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CIRCLE

The equation of a circle with center (h, k) and radius r is

This equation is also called the standard form of an equation of a circle with radius r and center (h, k).

x h 2 y k 2 r2 .

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EXAMPLE 7 Finding the Equation of a Circle

Find the center–radius form of the equation of the circle with center (–3, 4) and radius 7.

x h 2 y k 2 r2

x 3 2 y 4 2 72

x 3 2 y 4 2 49

Solution

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EXAMPLE 8 Graphing a Circle

Graph each equation.

b. x 2 2 y 3 2 25a. x2 y2 1

Solution

a. x2 y2 1

x 0 2 y 0 2 12

Center: (0, 0) Radius: 1 Called the unit circle

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EXAMPLE 8 Graphing a Circle

Solution continued

Center: (–2, 3) Radius: 5

b. x 2 2 y 3 2 25

x 2 2 y 3 2 52

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EQUATION CIRCLENote that stating that the equation

x 3 2 y 4 2 25

represents the circle of radius 5 with center (–3, 4) means two things:

(i) If the values of x and y are a pair of numbers that satisfy the equation, then they are the coordinates of a point on the circle with radius 5 and center (–3, 4).

(ii) If a point is on the circle, then its coordinates satisfy the equation.

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GENERAL FORM OF THE EQUATION OF A CIRCLE

The general form of the equation of a circle is

x2 y2 ax by c 0.

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EXAMPLE 9Converting the General Form toCenter-Radius Form

Find the center and radius of the circle with equation x2 y2 6x 8y 10 0.

Solution

Complete the squares on both x and y.

x2 6x y2 8y 10

x2 6x 9 y2 8y 16 10 9 16

x 3 2 y 4 215

Center: (3, – 4) Radius: 15 3.9