Graphs Of Equations; Circles

Chapter 1

7

Graphs

1.2 Graphs of Equations; Circles

1. add, 4 3. intercepts

5. (3,–4) 7. True

9. 11.

13. 15.

17.

19. (a) (–1, 0), (1, 0) (b) symmetric with respect to the x-axis,

y-axis, and origin

Chapter 1 Graphs

8

21. (a) -

p

2

,0

æ

è

ç

ö

ø

÷ ,

p

2

, 0

æ

è

ç

ö

ø

÷ , (0,1) (b) symmetric with respect to the y-axis

23. (a) (0, 0) (b) symmetric with respect to the x-axis

25. (a) (1, 0) (b) not symmetric with respect to x-axis,

y-axis, or origin

27. (a) (–1, 0), (1, 0), (0, –1) (b) symmetric with respect to the y-axis

29. (a) none (b) symmetric with respect to the origin

31. y = x 4 - x

0 = 04 - 0

0 = 0

1=14 - 1

1¹ 0

0 = (-1)4 - -1

0 ¹1- -1

(0, 0) is on the graph of the equation.

33. y 2 = x2 + 9

32 = 02 + 9

9 = 9

02 = 32 + 9

0 ¹18

02 = (-3)2 + 9

0 ¹18

(0, 3) is on the graph of the equation.

35. x 2 + y 2 = 4

02 + 22 = 4

4 = 4

(-2)2 + 22 = 4

8 ¹ 4

( 2)2

+ ( 2)2

= 4

4 = 4

(0, 2) and ( 2, 2) are on the graph of the equation.

37. x2 = y

y - intercept : Let x = 0, then 0 2 = yÞ y = 0 (0,0)

x - intercept : Let y = 0, then x2 = 0Þ x = 0 (0,0)

Test for symmetry:

x - axis : Replace y by - y : x2 = -y, which is not equivalent to x 2 = y.

y - axis : Replace x by - x : (-x)2 = y or x2 = y, which is equivalent to x 2 = y.

Origin : Replace x by - x and y by - y : (-x)2 = -y or x2 = -y,

which is not equivalent to x 2 = y.

Therefore, the graph is symmetric with respect to the y - axis .

39. y = 3x

y - intercept : Let x = 0, then y = 3× 0 = 0 (0,0)

x - intercept : Let y = 0, then 3 x = 0Þ x = 0 (0,0)

Section 1.2 Graphs of Equations; Circles

9

Test for symmetry:

x - axis : Replace y by - y : - y = 3x, which is not equivalent to y = 3x.

y - axis : Replace x by - x : y = 3(-x ) or y = -3x,

which is not equivalent to y = 3x.

Origin : Replace x by - x and y by - y : - y = 3(-x) or y = 3x,

which is equivalent to y = 3x.

Therefore, the graph is symmetric with respect to the origin.

41. x 2 + y - 9 = 0

y - intercept : Let x = 0, then 0 + y -9 = 0 Þ y = 9 (0,9)

x - intercept : Let y = 0, then x2 - 9 = 0Þ x = ±3 (-3,0),(3,0)

Test for symmetry:

x - axis : Replace y by - y : x2 + (-y)- 9 = 0 or x2 - y - 9 = 0,

which is not equivalent to x2 + y -9 = 0.

y - axis : Replace x by - x : (-x)2 + y - 9 = 0 or x2 + y -9 = 0,

which is equivalent to x 2 + y -9 = 0.

Origin : Replace x by - x and y by - y : (-x)2 + (-y)-9 = 0 or x 2 - y -9 = 0,

which is not equivalent to x 2 + y - 9 = 0.

Therefore, the graph is symmetric with respect to the y-axis.

43. 9 x2 + 4 y2 = 36

y - intercept : Let x = 0, then 4y2 = 36Þ y 2 = 9Þ y = ±3 (0,-3),(0,3)

x - intercept : Let y = 0, then 9x 2 = 36Þ x 2 = 4Þ x = ±2 (-2,0),(2,0)

Test for symmetry:

x - axis : Replace y by - y : 9x2 + 4(-y)2 = 36 or 9x 2 + 4y 2 = 36,

which is equivalent to 9x 2 + 4y 2 = 36.

y - axis : Replace x by - x : 9(-x )2 + 4y2 = 36 or 9x 2 + 4y 2 = 36,

which is equivalent to 9x 2 + 4y 2 = 36.

Origin : Replace x by - x and y by - y : 9(-x)2 + 4(-y)2 = 36 or 9x2 + 4y2 = 36,

which is equivalent to 9x 2 + 4y 2 = 36.

Therefore, the graph is symmetric with respect to the x-axis, the y-axis, and the origin.

45. y = x3 - 27

y - intercept : Let x = 0, then y = 03 - 27Þ y = -27 (0,-27)

x - intercept : Let y = 0, then 0 = x3 -27 Þ x 3 = 27 Þ x = 3 (3,0)

Chapter 1 Graphs

10

Test for symmetry:

x - axis : Replace y by - y : - y = x 3 -27, which is not equivalent to y = x 3 - 27.

y - axis : Replace x by - x : y = (-x)3 -27 or y = -x3 -27,

which is not equivalent to y = x 3 -27.

Origin : Replace x by - x and y by - y : - y = (-x)3 -27 or

y = x 3 + 27, which is not equivalent to y = x 3 -27.

Therefore, the graph is not symmetric with respect to the x-axis, the y-axis, or the origin.

47. y = x2 - 3x - 4

y - intercept : Let x = 0, then y = 02 - 3(0) - 4 Þ y = -4 (0,-4)

x - intercept : Let y = 0, then 0 = x 2 - 3x - 4 Þ(x - 4)(x +1) = 0Þ x = 4, x = -1 (4,0), (-1,0)

Test for symmetry:

x - axis : Replace y by - y : - y = x 2 - 3x - 4, which is not

equivalent to y = x 2 - 3x - 4.

y - axis : Replace x by - x : y = (-x)2 - 3(-x) - 4 or y = x 2 + 3x - 4,

which is not equivalent to y = x 2 - 3x - 4.

Origin : Replace x by - x and y by - y : - y = (-x)2 - 3(-x) - 4 or

y = -x 2 - 3x + 4, which is not equivalent to y = x 2 - 3x - 4.

Therefore, the graph is not symmetric with respect to the x-axis, the y-axis, or the origin.

49. y =

3x

x 2 + 9

y - intercept : Let x = 0, then y =

0

0 + 9

= 0 (0, 0)

x - intercept : Let y = 0, then 0 =

3x

x 2 + 9

Þ 3x = 0Þ x = 0 (0, 0)

Test for symmetry:

x - axis : Replace y by - y : - y =

3x

x2 + 9

, which is not

equivalent to y = 3x

x2 + 9

.

y - axis : Replace x by - x : y =

3(-x)

(-x)2 + 9

or y =

-3x

x2 + 9

,

which is not equivalent to y =

3x

x2 + 9

.

Origin : Replace x by - x and y by - y : - y =

-3x

(-x)2 + 9

or

y =

3x

x2 + 9

, which is equivalent to y =

3x

x 2 + 9

.

Therefore, the graph is symmetric with respect to the origin.

Section 1.2 Graphs of Equations; Circles

11

51. y = -x3

x 2 - 9

y - intercept : Let x = 0, then y =

0

-9

= 0 (0,0)

x - intercept : Let y = 0, then 0 =

-x 3

x2 -9

Þ-x 3 = 0 Þ x = 0 (0,0)

Test for symmetry:

x - axis : Replace y by - y : - y = -x 3

x 2 - 9

, which is not equivalent to y = -x3

x 2 - 9

.

y - axis : Replace x by - x : y =

-(-x)3

(-x)2 - 9

or y =

x 3

...