Intermediate Algebra (11th edition)

646 CHAPTER 11 Nonlinear Functions, Conic Sections, and Nonlinear Systems

Graphing an Ellipse Shifted Horizontally and Vertically

Graph

Just as and would indi-

cate that the center of a circle would be

so it is with this ellipse. FIGURE 18shows that the

graph goes through the four points

and

The x-values of these points are found by adding

to 2, and the y-values come from

adding to b= 7 - 3.

a= 5

1 - 3, - 32.

1 2, 4 2 , 1 7, - 32 , 1 2, - 102 ,

1 2, - 32 ,

1 x- 222 1 y+ 322

1 x- 222

25

+

1 y+ 322

49

=1.

EXAMPLE 6

x

y

–3 0

–10

4

7 (2, –3)

(x – 2)^2 25

(y + 3)^2 += 1 49

Add 7.

Add –7.

Add –5. Add 5.

FIGURE 18

A graphing calculator in function mode cannot directly graph a circle or an el-

lipse, since they do not represent functions. We must first solve the equation for y,

getting two functions and The union of these two graphs is the graph of the en-

tire figure.

For example, to graph begin by solving for y.

Subtract

Take square roots.

y=- 2 225 - 1 x+ 322 Add .- 2

y+ 2 = 225 - 1 x+ 322

1 y+ 222 = 25 - 1 x+ 322 1 x+ 322.

1 x+ 322 + 1 y+ 222 = 25

1 x+ 322 + 1 y+ 222 =25,

y 1 y 2.

CONNECTIONS

NOW TRY

NOTE Graphs of circles and ellipses are not graphs of functions.The only conic

section whose graph represents a function is the vertical parabola with equation

ƒ 1 x 2 =ax^2 + bx+c.

The two functions to be graphed are

and

To get an undistorted screen, a square viewing windowmust be used. (Refer to your

instruction manual for details.) See FIGURE 19. The two semicircles seem to be

disconnected. This is because the graphs are nearly vertical at those points, and the

calculator cannot show a true picture of the behavior there.

For Discussion or Writing

Find the two functions and to use to obtain the graph of the circle with equa-

tion Then graph the circle using a square viewing

window.

1 x- 322 + 1 y+ 122 =36.

y 1 y 2

y 1 =- 2 + 225 - 1 x+ 322 y 2 =- 2 - 225 - 1 x+ 322.

Remember bothroots.

10