ch. 2 graphs...f - ; - f + ; c

Ch. 2 Graphs

2.1 The Distance and Midpoint Formulas

1 Rectangular Coordinates

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Name the quadrant in which the point is located.1) (20, 5)

A) I B) II C) III D) IV

2) (-13, 5)A) I B) II C) III D) IV

3) (-13, -4)A) I B) II C) III D) IV

4) (6, -2)A) I B) II C) III D) IV

Identify the points in the graph for the ordered pairs.

x-5 5

y

5

-5

AB

C

D

E

FG

H

IJ K

L

x-5 5

y

5

-5

AB

C

D

E

FG

H

IJ K

L

5) (0, 2), (4, 3)A) F and E B) B and C C) C and E D) C and K

6) (-5, -4), (0, -3)A) G and I B) A and J C) I and J D) A and G

7) (-3, 4), (2, 0), (4, -5)A) A, B, and F B) F, K, and L C) B, F, and L D) B, C, and L

8) (3, 5), (-3, 0)A) I and G B) D and G C) D and J D) L and J

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Give the coordinates of the points shown on the graph.9)

x-5 5

y

5

-5

A

B

x-5 5

y

5

-5

A

B

A) A = (2, 28), B = (4, -3) B) A = (6, 2), B = (4, -3)C) A = (6, 4), B = (2, 4) D) A = (6, 2), B = (-3, 4)

10)

x-5 5

y

5

-5

C

Dx-5 5

y

5

-5

C

D

A) C = (-4, -3), D = (3, -3) B) C = (-4, 3), D = (2, -3)C) C = (3, -4), D = (-3, 2) D) C = (-4, 3), D = (-3, 2)

11)

x-5 5

y

5

-5

E

Fx-5 5

y

5

-5

E

F

A) E = (-7, -3) , F = (-2, 2) B) E = (-2, -3), F = (2, -3)C) E = (2, -2), F = (-3, -7) D) E = (-2, 2), F = (-7, -3)


12)

x-5 5

y

5

-5 G

H

x-5 5

y

5

-5 G

H

A) G = (4, -6), H = (7, -2) B) G = (4, 7), H = (-6, 7)C) G = (4, -6), H = (-2, 7) D) G = (-6, 4), H = (7, -2)

Plot the point in the xy-plane. Tell in which quadrant or on what axis the point lies.13) (3, 1)

x-5 5

y

5

-5

x-5 5

y

5

-5

A)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant IV

B)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant I


C)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant I

D)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant II

14) (-4, 1)

x-5 5

y

5

-5

x-5 5

y

5

-5

A)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant I

B)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant IV


C)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant II

D)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant III

15) (5, -3)

x-5 5

y

5

-5

x-5 5

y

5

-5

A)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant I

B)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant IV


C)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant III

D)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant II

16) (-3, -1)

x-5 5

y

5

-5

x-5 5

y

5

-5

A)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant II

B)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant III


C)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant IV

D)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant III

17) (0, -5)

x-5 5

y

5

-5

x-5 5

y

5

-5

A)

x-5 5

y

5

-5

x-5 5

y

5

-5

y-axis

B)

x-5 5

y

5

-5

x-5 5

y

5

-5

x-axis


C)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant II

D)

x-5 5

y

5

-5

x-5 5

y

5

-5

y-axis

18) (3, 0)

x-5 5

y

5

-5

x-5 5

y

5

-5

A)

x-5 5

y

5

-5

x-5 5

y

5

-5

x-axis

B)

x-5 5

y

5

-5

x-5 5

y

5

-5

x-axis


C)

x-5 5

y

5

-5

x-5 5

y

5

-5

y-axis

D)

x-5 5

y

5

-5

x-5 5

y

5

-5

Quadrant II

2 Use the Distance Formula


Find the distance d(P1, P2) between the points P1 and P2.1)

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

A) 6 B) 5 C) 7 D) 26

2)

x-8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

x-8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

A) 56 B) 1 C) 15 D) 113


3)

x-8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

x-8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

A) 2 13 B) 20 C) 20 5 D) 10

4)

x-8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

x-8 -6 -4 -2 2 4 6 8

y8

6

4

2

-2

-4

-6

-8

A) 140 35 B) 10 C) 2 37 D) 140

5) P1 = (-3, -3); P2 = (-3, 4)

A) 8 B) 6 C) 7 D) 7

6) P1 = (-3, 3); P2 = (-15, -2)A) 13 B) 14 C) 26 D) 169

7) P1 = (0, -10); P2 = (-5, -10)

A) 25 B) 5 5 C) 10 D) 5

8) P1 = (0, 0); P2 = (5, 9)

A) 106 B) 14 C) 3 5 D) 106

9) P1 = (7, 2); P2 = (-1, -7)

A) 1 B) 17 C) 145 D) 72

10) P1 = (3, -5); P2 = (5, -1)

A) 12 B) 12 3 C) 2 D) 2 5


11) P1 = (-5, -5); P2 = (3, 7)

A) 4 B) 4 13 C) 80 D) 80 5

12) P1 = (-0.5, 0.8); P2 = (-2.7, 1.2) Round to three decimal places, if necessary.A) 7.071 B) 13 C) 2.236 D) 2.336

Decide whether or not the points are the vertices of a right triangle.13) (-1, 5), (1, 5), (1, 9)

A) Yes B) No

14) (5, 13), (7, 17), (9, 16)A) Yes B) No

15) (-4, 7), (2, 9), (1, 4)A) Yes B) No

16) (6, -7), (12, -5), (18, -12)A) Yes B) No

Solve the problem.17) Find all values of k so that the given points are 29 units apart.

(-5, 5), (k, 0)A) -3, -7 B) 3, 7 C) -7 D) 7

18) Find the area of the right triangle ABC with A = (-2, 7), B = (7, -1), C = (3, 9).

A) 29 square units B) 582 square units C) 29

2 square units D) 58 square units

19) Find all the points having an x-coordinate of 9 whose distance from the point (3, -2) is 10.A) (9, 2), (9, -4) B) (9, 6), (9, -10) C) (9, -12), (9, 8) D) (9, 13), (9, -7)

20) A middle schoolʹs baseball playing field is a square, 55 feet on a side. How far is it directly from home plate tosecond base (the diagonal of the square)? If necessary, round to the nearest foot.

A) 79 feet B) 77 feet C) 85 feet D) 78 feet

21) A motorcycle and a car leave an intersection at the same time. The motorcycle heads north at an average speedof 20 miles per hour, while the car heads east at an average speed of 48 miles per hour. Find an expression fortheir distance apart in miles at the end of t hours.

A) t 68 miles B) 52t miles C) 52 t miles D) 2t 13 miles

22) A rectangular city park has a jogging loop that goes along a length, width, and diagonal of the park. To thenearest yard, find the length of the jogging loop, if the length of the park is 125 yards and its width is 75 yards.

A) 345 yards B) 346 yards C) 146 yards D) 145 yards


23) Find the length of each side of the triangle determined by the three points P1, P2, and P3. State whether thetriangle is an isosceles triangle, a right triangle, neither of these, or both.P1 = (-5, -4), P2 = (-3, 4), P3 = (0, -1)

A) d(P1, P2) = 2 17; d(P2, P3) = 34; d(P1, P3) = 34isosceles triangle

B) d(P1, P2) = 2 17; d(P2, P3) = 34; d(P1, P3) = 34both

C) d(P1, P2) = 2 17; d(P2, P3) = 34; d(P1, P3) = 5 2neither

D) d(P1, P2) = 2 17; d(P2, P3) = 34; d(P1, P3) = 5 2right triangle

3 Use the Midpoint Formula


Find the midpoint of the line segment joining the points P1 and P2.1) P1 = (1, 3); P2 = (6, 8)

A) (-5, -5) B) (72, 112) C) (7, 11) D) (11

2, 72)

2) P1 = (-5, 4); P2 = (9, 7)

A) 4, 11 B) 2, 112

C) -14, -3 D) - 7, - 32

3) P1 = (7, 1); P2 = (-16, -16)

A) - 92, - 15

2B) -9, -15 C) 9, 15 D) 23

2, 172

4) P1 = (-0.4, 0.5); P2 = (2.2, -2.6)A) (0.9, -1.05) B) (-1.55, 1.3) C) (-1.05, 0.9) D) (1.3, -1.55)

5) P1 = (x, 8); P2 = (0, 9)

A) x, 17 B) - x2, 1 C) x, 17

2D) x

2, 172

6) P1 = (9b, 7); P2 = (10b, 2)

A) b, 5 B) 9b2, 192

C) 19b2

, 92

D) 19b, 9

Solve the problem.7) If (2, -5) is the endpoint of a line segment, and (4, -2) is its midpoint, find the other endpoint.

A) (-2, -11) B) (8, -1) C) (6, 1) D) (6, -8)

8) If (-5, 3) is the endpoint of a line segment, and (-6, -1) is its midpoint, find the other endpoint.A) (-7, -5) B) (-13, 1) C) (-3, 11) D) (-7, 7)

9) If (-6, 4) is the endpoint of a line segment, and (-2, -1) is its midpoint, find the other endpoint.A) (2, -6) B) (2, 9) C) (-16, 12) D) (-14, 14)


10) If (-3, 4) is the endpoint of a line segment, and (-8, 8) is its midpoint, find the other endpoint.A) (-13, 12) B) (-13, 0) C) (5, -6) D) (7, -4)

11) The medians of a triangle intersect at a point. The distance from the vertex to the point is exactly two -thirds ofthe distance from the vertex to the midpoint of the opposite side. Find the exact distance of that point from thevertex A(3, 4) of a triangle, given that the other two vertices are at (0, 0) and (8, 0).

A) 173

B) 83

C) 2 173

D) 2

2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry

1 Graph Equations by Plotting Points


Determine whether the given point is on the graph of the equation.1) Equation: y = x3 - x

Point: (4, 62)A) Yes B) No

2) Equation: x2 + y2 = 4Point: ( 2, 2)

A) No B) Yes


Graph the equation by plotting points.3) y = x + 5

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


4) y = 3x - 9

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


5) y = x2 + 1

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


6) 5x + 3y = 15

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


7) x2 + 4y = 16

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

Solve the problem.8) If (a, 3) is a point on the graph of y = 2x - 5, what is a?

A) -4 B) 1 C) 4 D) -1

9) If (3, b) is a point on the graph of 3x - 2y = 17, what is b?

A) 113

B) 4 C) 233

D) -4

10) The height of a baseball (in feet) at time t (in seconds) is given by y = -16x2 + 80x + 5. Which one of thefollowing points is not on the graph of the equation?

A) (2, 117) B) (4, 69) C) (1, 69) D) (3, 101)


2 Find Intercepts from a Graph


List the intercepts of the graph.1)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) (-2, 0), (0, 2) B) (0, -2), (0, 2) C) (0, -2), (2, 0) D) (-2, 0), (2, 0)

2)

x-5 5

y

5

-5

x-5 5

y

5

-5

A) (-1, 0) B) (-1, -1) C) (0, 0) D) (0, -1)


3)

x-π -π

2π2 π

y54321

-1-2-3-4-5

x-π -π

2π2 π

y54321

-1-2-3-4-5

A) - π2, 0 , (3, 0), π

2, 0 B) - π

2, 0 , (0, 3), π

2, 0

C) 0, - π2

, (3, 0), 0, π2

D) 0, - π2

, (0, 3), 0, π2

4)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) (0, -1), (0, 3), (3, 0) B) (-1, 0), (0, 3), (0, 3) C) (0, -1), (3, 0), (0, 3) D) (-1, 0), (0, 3), (3, 0)

5)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) (4, 0) B) (0, -4) C) (-4, 0) D) (0, 4)


6)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) (-5, 0), (0, -5), (0, 5), (5, 0) B) (0, 5), (5, 0)C) (-5, 0), (0, -5), (0, 0), (0, 5), (5, 0) D) (-5, 0), (0, 5)

7)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) (-2, 0), (1, 0) (-5, 0), (0, -2) B) (2, 0), (1, 0), (5, 0), (0, -2)C) (-2, 0), (0, -2), (0, 1), (0, -5) D) (-2, 0), (0, 2), (0, 1), (0, 5)

8)

A) (-4, 0), (0, 4), (4, 0) B) (-2, 0), (0, 4), (2, 0) C) (-2, 0), (0, 2), (2, 0) D) (-2, 0), (2, 0)


3 Find Intercepts from an Equation


List the intercepts for the graph of the equation.1) y = x + 1

A) (1, 0), (0, -1) B) (-1, 0), (0, -1) C) (1, 0), (0, 1) D) (-1, 0), (0, 1)

2) y = 3xA) (0, 0) B) (3, 0) C) (0, 3) D) (3, 3)

3) y2 = x + 1A) (0, -1), (-1, 0), (0, 1) B) (-1, 0), (0, -1), (1, 0)C) (1, 0), (0, 1), (0, -1) D) (0, -1), (1, 0), (0, 1)

4) y = 8x

A) (1, 0) B) (1, 1) C) (0, 1) D) (0, 0)

5) x2 + y - 81 = 0A) (-9, 0), (0, -81), (9, 0) B) (0, -9), (81, 0), (0, 9)C) (-9, 0), (0, 81), (9, 0) D) (9, 0), (0, 81), (0, -81)

6) 4x2 + 16y2 = 64A) (-4, 0), (0, -2), (0, 2), (4, 0) B) (-4, 0), (-16, 0), (16, 0), (4, 0)C) (-16, 0), (0, -4), (0, 4), (16, 0) D) (-2, 0), (-4, 0), (4, 0), (2, 0)

7) 9x2 + y2 = 9A) (-9, 0), (0, -1), (0, 1), (9, 0) B) (-1, 0), (0, -3), (0, 3), (1, 0)C) (-3, 0), (0, -1), (0, 1), (3, 0) D) (-1, 0), (0, -9), (0, 9), (1, 0)

8) y = x3 - 27A) (-27, 0), (0, 3) B) (0, -3), (0, 3) C) (0, -3), (-3, 0) D) (0, -27), (3, 0)

9) y = x4 - 16A) (0, 16), (-2, 0), (2, 0) B) (0, -16)C) (0, 16) D) (0, -16), (-2, 0), (2, 0)

10) y = x2 + 13x + 36A) (0, -9), (0, -4), (36, 0) B) (0, 9), (0, 4), (36, 0)C) (9, 0), (4, 0), (0, 36) D) (-9, 0), (-4, 0), (0, 36)

11) y = x2 + 4A) (0, 4), (-2, 0), (2, 0) B) (0, 4) C) (4, 0) D) (4, 0), (0, -2), (0, 2)

12) y = 7xx2 + 49

A) (-49, 0), (0, 0), (49, 0) B) (0, 0)C) (0, -7), (0, 0), (0, 7) D) (-7, 0), (0, 0), (7, 0)


13) y = x2 - 497x4

A) (-49, 0), (0, 0), (49, 0) B) (0, -7), (0, 7)C) (-7, 0), (7, 0) D) (0, 0)


4 Test an Equation for Symmetry with Respect to the x-Axis, the y-Axis, and the Origin


Plot the point A. Plot the point B that has the given symmetry with point A.1) A = (-3, 1); B is symmetric to A with respect to the y-axis

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A

B

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A

B

B)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A

B

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A

B

C)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A

B

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A

B

D)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A

B

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A

B


2) A = (0, 5); B is symmetric to A with respect to the origin

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

AB

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

AB

B)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A

B

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A

B

C)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A

B

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A

B

D)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A

B

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

A

B


List the intercepts of the graph.Tell whether the graph is symmetric with respect to the x -axis, y-axis, origin, or none ofthese.

3)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) intercepts: (-6, 0) and (6, 0)symmetric with respect to origin

B) intercepts: (-6, 0) and (6, 0)symmetric with respect to x-axis, y-axis, and origin

C) intercepts: (0, -6) and (0, 6)symmetric with respect to x-axis, y-axis, and origin

D) intercepts: (0, -6) and (0, 6)symmetric with respect to y-axis

4)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) intercepts: (6, 0) and (-6, 0symmetric with respect to y-axis

B) intercepts: (0, 6) and (0, -6)symmetric with respect to x-axis, y-axis, and origin

C) intercepts: (0, 6) and (0, -6)symmetric with respect to origin

D) intercepts: (6, 0) and (-6, 0)symmetric with respect to x-axis, y-axis, and origin


5)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) intercept: (10, 0)symmetric with respect to y-axis

B) intercept: (0, 10)no symmetry

C) intercept: (0, 10)symmetric with respect to x-axis

D) intercept: (10, 0)no symmetry

6)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


B) intercept: (0, 3)symmetric with respect to origin

C) intercept: (3, 0)symmetric with respect to x-axis

D) intercept: (3, 0)symmetric with respect to y-axis


7)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) intercepts: (-3, 0), (0, 0), (3, 0)symmetric with respect to x-axis, y-axis, and origin

B) intercepts: (-3, 0), (0, 0), (3, 0)symmetric with respect to y-axis

C) intercepts: (-3, 0), (0, 0), (3, 0)symmetric with respect to origin

D) intercepts: (-3, 0), (0, 0), (3, 0)symmetric with respect to x-axis

Draw a complete graph so that it has the given type of symmetry.8) Symmetric with respect to the y-axis

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

(0, 3)

(2, -1) x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

(0, 3)

(2, -1)

A)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

B)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5


C)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

D)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

9) origin

x-π -π

2π2 π

y5

4

3

2

1

-1

-2

-3

-4

-5

x-π -π

2π2 π

y5

4

3

2

1

-1

-2

-3

-4

-5

A)

x-π -π

2π2 π

y5

4

3

2

1

-1

-2

-3

-4

-5

x-π -π

2π2 π

y5

4

3

2

1

-1

-2

-3

-4

-5

B)

x-π -π

2π2 π

y5

4

3

2

1

-1

-2

-3

-4

-5

x-π -π

2π2 π

y5

4

3

2

1

-1

-2

-3

-4

-5


C)

x-π -π

2π2 π

y5

4

3

2

1

-1

-2

-3

-4

-5

x-π -π

2π2 π

y5

4

3

2

1

-1

-2

-3

-4

-5

D)

x-π -π

2π2 π

y5

4

3

2

1

-1

-2

-3

-4

-5

x-π -π

2π2 π

y5

4

3

2

1

-1

-2

-3

-4

-5

10) Symmetric with respect to the x-axis

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

(2, 0)

(3, 1)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

(2, 0)

(3, 1)

A)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

B)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5


C)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

D)

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

x-5 -4 -3 -2 -1 1 2 3 4 5

y5

4

3

2

1

-1

-2

-3

-4

-5

List the intercepts and type(s) of symmetry, if any.11) y2 = x + 1

A) intercepts: (1, 0), (0, 1), (0, -1)symmetric with respect to x-axis

B) intercepts: (-1, 0), (0, 1), (0, -1)symmetric with respect to x-axis

C) intercepts: (0, 1), (1, 0), (-1, 0)symmetric with respect to y-axis

D) intercepts: (0, -1), (1, 0), (-1, 0)symmetric with respect to y-axis

12) 4x2 + 9y2 = 36A) intercepts: (2, 0), (-2, 0), (0, 3), (0, -3)

symmetric with respect to the originB) intercepts: (3, 0), (-3, 0), (0, 2), (0, -2)

symmetric with respect to x-axis and y-axisC) intercepts: (3, 0), (-3, 0), (0, 2), (0, -2)

symmetric with respect to x-axis, y-axis, and originD) intercepts: (2, 0), (-2, 0), (0, 3), (0, -3)

symmetric with respect to x-axis and y-axis

13) y = -xx2 - 3


B) intercept: (0, 0)symmetric with respect to x-axis

C) intercept: (0, 0)symmetric with respect to origin

D) intercepts: ( 3, 0), (- 3, 0), (0, 0)symmetric with respect to origin

Determine whether the graph of the equation is symmetric with respect to the x -axis, the y-axis, and/or the origin.14) y = x + 1

A) x-axisB) originC) y-axisD) x-axis, y-axis, originE) none


15) y = -3xA) y-axisB) x-axisC) originD) x-axis, y-axis, originE) none

16) x2 + y - 25 = 0A) x-axisB) y-axisC) originD) x-axis, y-axis, originE) none

17) y2 - x - 81 = 0A) originB) y-axisC) x-axisD) x-axis, y-axis, originE) none

18) 4x2 + 16y2 = 64A) y-axisB) x-axisC) originD) x-axis, y-axis, originE) none

19) 4x2 + y2 = 4A) originB) y-axisC) x-axisD) x-axis, y-axis, originE) none

20) y = x2 + 7x + 10A) originB) x-axisC) y-axisD) x-axis, y-axis, originE) none

21) y = 9xx2 + 81

A) x-axisB) y-axisC) originD) x-axis, y-axis, originE) none


22) y = x2 - 648x4

A) originB) x-axisC) y-axisD) x-axis, y-axis, originE) none

23) y = 4x2 + 2A) x-axisB) y-axisC) originD) x-axis, y-axis, originE) none

24) y = (x - 4)(x - 2)A) originB) x-axisC) y-axisD) x-axis, y-axis, originE) none

25) y = -6x3 + 3xA) y-axisB) x-axisC) originD) x-axis, y-axis, originE) none

26) y = -8x4 + 3x - 1A) originB) y-axisC) x-axisD) x-axis, y-axis, originE) none

Solve the problem.27) If a graph is symmetric with respect to the y-axis and it contains the point (5, -6), which of the following points

is also on the graph?A) (-5, 6) B) (-6, 5) C) (-5, -6) D) (5, -6)

28) If a graph is symmetric with respect to the origin and it contains the point (-4, 7), which of the following pointsis also on the graph?

A) (7, -4) B) (4, 7) C) (4, -7) D) (-4, -7)


5 Know How to Graph Key Equations


Graph the equation by plotting points.1) y = x3

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


2) x = y2

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


3) y = x

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


4) y = 1x

x-5 5

y

5

-5

x-5 5

y

5

-5

A)

x-5 5

y

5

-5

x-5 5

y

5

-5

B)

x-5 5

y

5

-5

x-5 5

y

5

-5

C)

x-5 5

y

5

-5

x-5 5

y

5

-5

D)

x-5 5

y

5

-5

x-5 5

y

5

-5


2.3 Lines

1 Calculate and Interpret the Slope of a Line


Find the slope of the line through the points and interpret the slope.1)

x-10 -5 5 10

y10

5

-5

-10

(3, 1)(0, 0)

x-10 -5 5 10

y10

5

-5

-10

(3, 1)(0, 0)

A) 13; for every 3-unit increase in x, y will increase by 1 unit

B) - 13; for every 3-unit increase in x, y will decrease by 1 unit

C) 3; for every 1-unit increase in x, y will increase by 3 unitsD) -3; for every 1-unit increase in x, y will decrease by 3 units

Find the slope of the line.2)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) - 13

B) 13

C) - 3 D) 3


3)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) -1 B) 5 C) 1 D) -5

4)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) -4 B) 4 C) -1 D) 1

5)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) - 12

B) 12

C) -2 D) 2

Find the slope of the line containing the two points.6) (7, -6); (-5, 7)

A) - 1213

B) 1312

C) - 1312

D) 1213


7) (9, 0); (0, 4)

A) 94

B) 49

C) - 94

D) - 49

8) (7, 6); (8, 8)

A) 2 B) - 2 C) - 12

D) 12

9) (-4, -7); (-4, -8)A) 1 B) 0 C) -1 D) undefined

10) (2, 7); (-1, 7)

A) 13

B) 0 C) -3 D) undefined


2 Graph Lines Given a Point and the Slope


Graph the line containing the point P and having slope m.

1) P = (7, -3); m = - 1011

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


2) P = (-3, -9); m = 1

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


3) P = (-2, 0); m = - 43

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


4) P = (0, 3); m = 43

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


5) P = (0, 4); m = - 45

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


6) P = (-2, 0); m = 1

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


7) P = (2, 0); m = - 43

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


8) P = (-7, -1); m = 0

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


9) P = (9, -3); slope undefined

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

3 Find the Equation of a Vertical Line


Find an equation for the line with the given properties.1) Slope undefined; containing the point (7, -8)

A) x = -8 B) x = 7 C) y = 7 D) y = -8

2) Vertical line; containing the point (-2, -1)A) x = -2 B) x = -1 C) y = -1 D) y = -2


3) Slope undefined; containing the point - 57, 1

A) y = - 57

B) y = 1 C) x = - 57

D) x = 1

4) Vertical line; containing the point (6.3, -1.5)A) x = -1.5 B) x = 4.8 C) x = 6.3 D) x = 0

4 Use the Point-Slope Form of a Line; Identify Horizontal Lines


Find the slope-intercept form of the equation of the line with the given properties.1) Horizontal; containing the point (-5, -6)

A) y = -5 B) x = -5 C) y = -6 D) x = -6

2) Slope = 0; containing the point (3, 5)A) x = 3 B) y = 3 C) x = 5 D) y = 5

3) Horizontal; containing the point - 49, 5

A) y = 0 B) y = - 49

C) y = 5 D) y = -5

4) Horizontal; containing the point (-5.4, -8.0)A) y = -8.0 B) y = -5.4 C) y = 0 D) y = 13.4


Find the slope of the line and sketch its graph.5) y + 1 = 0

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) slope = - 1

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B) slope is undefined

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C) slope = -1

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D) slope = 0

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


5 Find the Equation of a Line Given Two Points


Find the equation of the line in slope-intercept form.1)

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

x-6 -4 -2 2 4 6

y6

4

2

-2

-4

-6

A) y = - 3x + 9 B) y = - 3x - 7 C) y = - 3x + 11 D) y = - 13x - 1

7

Find an equation for the line, in the indicated form, with the given properties.2) Containing the points (-7, 6) and (4, -7); slope-intercept form

A) y - 6 = - 1311

(x + 7) B) y = 1311

x - 2511

C) y = - 1311

x - 2511

D) y = mx - 2511

3) Containing the points (3, -7) and (-5, 8); general formA) -10x + 13y = -54 B) 10x - 13y = -54 C) -15x + 8y = -11 D) 15x + 8y = -11

4) Containing the points (12, 0) and (0, -10); general form

A) y = - 56x - 10 B) 10x + 12y = 120 C) 10x - 12y = 120 D) y = - 5

6x + 12

5) Containing the points (-5, 8) and (4, -8); general formA) -13x + 12y = -44 B) 13x - 12y = -44 C) -16x - 9y = 8 D) 16x - 9y = 8

6) Containing the points (8, 3) and (0, -6); general formA) -5x - 6y = -36 B) 5x + 6y = -36 C) 9x + 8y = -48 D) -9x + 8y = -48

7) Containing the points (4, 0) and (2, 7); general formA) -4x + 5y = -43 B) -7x + 2y = 28 C) 4x - 5y = -43 D) 7x + 2y = 28

8) Containing the points (3, -5) and (1, 4); general formA) 8x - 3y = -20 B) 9x + 2y = 17 C) -8x + 3y = -20 D) -9x + 2y = 17


Solve.9) The relationship between Celsius (°C) and Fahrenheit (°F) degrees of measuring temperature is linear. Find an

equation relating °C and °F if 10°C corresponds to 50°F and 30°C corresponds to 86°F. Use the equation to findthe Celsius measure of 24° F.

A) C = 59F - 10; 10

3 °C B) C = 9

5F - 80; - 184

5 °C

C) C = 59F - 160

9; - 40

9 °C D) C = 5

9F + 160

9; 280

9 °C

10) A school has just purchased new computer equipment for $25,000.00. The graph shows the depreciation of theequipment over 5 years. The point (0, 25,000) represents the purchase price and the point (5, 0) represents whenthe equipment will be replaced. Write a linear equation in slope-intercept form that relates the value of theequipment, y, to years after purchase x . Use the equation to predict the value of the equipment after 3 years.

x2.5 5

y25000225002000017500150001250010000750050002500

x2.5 5

y25000225002000017500150001250010000750050002500

A) y = - 25,000x + 25,000;value after 3 years is $-50,000.00

B) y = 5000x - 25,000;value after 3 years is $10,000.00

C) y = - 5000x + 25,000;value after 3 years is $10,000.00;

D) y = 25,000x + 5;value after 3 years is $10,000.00

11) The average value of a certain type of automobile was $14,040 in 1991 and depreciated to $7800 in 1996. Let ybe the average value of the automobile in the year x, where x = 0 represents 1991. Write a linear equation thatrelates the average value of the automobile, y, to the year x.

A) y = - 11248

x - 7800 B) y = -1248x + 14,040

C) y = -1248x + 7800 D) y = -1248x + 1560

12) An investment is worth $3427 in 1991. By 1994 it has grown to $4444. Let y be the value of the investment inthe year x, where x = 0 represents 1991. Write a linear equation that relates the value of the investment, y, tothe year x.

A) y = 1339

x + 3427 B) y = -339x + 5461 C) y = 339x + 3427 D) y = -339x + 3427

13) A faucet is used to add water to a large bottle that already contained some water. After it has been filling for 3seconds, the gauge on the bottle indicates that it contains 8 ounces of water. After it has been filling for 11seconds, the gauge indicates the bottle contains 24 ounces of water. Let y be the amount of water in the bottle xseconds after the faucet was turned on. Write a linear equation that relates the amount of water in the bottle,y,to the time x.

A) y = 2x + 2 B) y = -2x + 14 C) y = 2x + 13 D) y = 12x + 13

2


14) When making a telephone call using a calling card, a call lasting 5 minutes cost $2.40. A call lasting 12 minutescost $5.20. Let y be the cost of making a call lasting x minutes using a calling card. Write a linear equation thatrelates the cost of a making a call, y, to the time x.

A) y = 0.4x - 6.8 B) y = 52x - 101

10C) y = -0.4x + 4.4 D) y = 0.4x + 0.4

15) A vendor has learned that, by pricing hot dogs at $1.25, sales will reach 124 hot dogs per day. Raising the priceto $2.00 will cause the sales to fall to 94 hot dogs per day. Let y be the number of hot dogs the vendor sells at xdollars each. Write a linear equation that relates the number of hot dogs sold per day, y, to the price x.

A) y = 40x + 74 B) y = -40x - 174 C) y = -40x + 174 D) y = - 140

x + 396732

16) A vendor has learned that, by pricing hot dogs at $1.25, sales will reach 93 hot dogs per day. Raising the priceto $2.25 will cause the sales to fall to 41 hot dogs per day. Let y be the number of hot dogs the vendor sells at xdollars each. Write a linear equation that relates the number of hot dogs sold per day to the price x.

A) y = 52x + 28 B) y = - 152

x + 19339208

C) y = -52x + 158 D) y = -52x - 158

6 Write the Equation of a Line in Slope-Intercept Form


Find the slope-intercept form of the equation of the line with the given properties.1) Slope = 5; containing the point (-2, -4)

A) y = -5x - 6 B) y = 5x - 6 C) y = 5x + 6 D) y = -5x + 6

2) Slope = 0; containing the point (5, -8)A) x = -8 B) x = 5 C) y = 5 D) y = -8

3) Slope = 6; y-intercept = 8A) y = 8x - 6 B) y = 6x - 8 C) y = 6x + 8 D) y = 8x + 6

4) x-intercept = 2; y-intercept = 7

A) y = - 72x + 7 B) y = - 7

2x + 2 C) y = - 2

7x + 2 D) y = 7

2x + 7

Write the equation in slope-intercept form.5) 5x + 7y = 13

A) y = 57x - 13

7B) y = 5x - 13 C) y = 5

7x + 13

7D) y = - 5

7x + 13

7

6) 6x + 7y = 5

A) y = 6x + 12 B) y = 76x - 5

6C) y = 6

7x + 5

7D) y = 12

7x + 5

7

7) 9x - 7y = 5

A) y = 97x + 5

7B) y = 7

9x + 5

9C) y = 9x - 5 D) y = 9

7x - 5

7

8) x = 7y + 2

A) y = 17x - 2

7B) y = 1

7x - 2 C) y = 7x - 2 D) y = x - 2

7


Solve.9) A truck rental company rents a moving truck one day by charging $25 plus $0.07 per mile. Write a linear

equation that relates the cost C, in dollars, of renting the truck to the number x of miles driven. What is the costof renting the truck if the truck is driven 130 miles?

A) C = 0.07x - 25; $15.90 B) C = 0.07x + 25; $25.91C) C = 0.07x + 25; $34.10 D) C = 25x + 0.07; $3250.07

10) Each week a soft drink machine sells x cans of soda for $0.75/soda. The cost to the owner of the soda machinefor each soda is $0.10. The weekly fixed cost for maintaining the soda machine is $25/week. Write an equationthat relates the weekly profit, P, in dollars to the number of cans sold each week. Then use the equation to findthe weekly profit when 92 cans of soda are sold in a week.

A) P = 0.65x + 25; $84.80 B) P = 0.65x - 25; $34.80C) P = 0.75x + 25; $94.00 D) P = 0.75x - 25; $44.00

11) Each day the commuter train transports x passengers to or from the city at $1.75/passenger. The daily fixed costfor running the train is $1200. Write an equation that relates the daily profit, P, in dollars to the number ofpassengers each day. Then use the equation to find the daily profit when the train has 920 passengers in a day.

A) P = 1200 - 1.75x; $410 B) P = 1.75x; $1610C) P = 1.75x + 1200; $2810 D) P = 1.75x - 1200; $410

12) Each month a beauty salon gives x manicures for $12.00/manicure. The cost to the owner of the beauty salon foreach manicure is $7.35. The monthly fixed cost to maintain a manicure station is $120.00. Write an equation thatrelates the monthly profit, in dollars, to the number of manicures given each month. Then use the equation tofind the monthly profit when 200 manicures are given in a month.

A) P = 4.65x - 120; $810 B) P =12x - 120; $2280C) P = 4.65x; $930 D) P = 7.35x - 120; $1350

13) Each month a gas station sells x gallons of gas at $1.92/gallon. The cost to the owner of the gas station for eachgallon of gas is $1.32. The monthly fixed cost for running the gas station is $37,000. Write an equation thatrelates the monthly profit, in dollars, to the number of gallons of gasoline sold. Then use the equation to findthe monthly profit when 75,000 gallons of gas are sold in a month.

A) P = 1.92x - 37,000; $107,000 B) P = 0.60x + 37,000; $82,000C) P = 1.32x - 37,000; $62,000 D) P = 0.60x - 37,000; $8000

7 Identify the Slope and y-Intercept of a Line from Its Equation


Find the slope and y-intercept of the line.

1) y = - 49x + 5

A) slope = 5; y-intercept = - 49

B) slope = - 94; y-intercept = - 5

C) slope = 49; y-intercept = - 5 D) slope = - 4

9; y-intercept = 5

2) x + y = 8A) slope = 0; y-intercept = 8 B) slope = 1; y-intercept = 8C) slope = -1; y-intercept = 8 D) slope = -1; y-intercept = -8


3) 7x + y = 12

A) slope = 7; y-intercept = 12 B) slope = - 17; y-intercept = 12

7

C) slope = -7; y-intercept = 12 D) slope = 712

; y-intercept = 112

4) -7x + 5y = 8

A) slope = 75; y-intercept = 8

5B) slope = 11

5; y-intercept = 8

5

C) slope = 57; y-intercept = - 8

7D) slope = 7; y-intercept = 11

5) 13x + 4y = 13


4B) slope = 13; y-intercept = 13

C) slope = 134; y-intercept = 13

4D) slope = - 13

4; y-intercept = 13

4

6) 9x - 7y = 4


7B) slope = 9; y-intercept = 4


9D) slope = 9

7; y-intercept = 4

7

7) 4x - 7y = 28

A) slope = - 47; y-intercept = 4 B) slope = 4; y-intercept = 28

C) slope = 74; y-intercept = 7 D) slope = 4

7; y-intercept = -4

8) x + 5y = 1

A) slope = 1; y-intercept = 1 B) slope = -5; y-intercept = 5


5D) slope = - 1

5; y-intercept = 1

5

9) -x + 12y = 72

A) slope = - 112

; y-intercept = 6 B) slope = -1; y-intercept = 72

C) slope = 12; y-intercept = -72 D) slope = 112

; y-intercept = 6

10) y = 9A) slope = 9; y-intercept = 0 B) slope = 1; y-intercept = 9C) slope = 0; no y-intercept D) slope = 0; y-intercept = 9

11) x = -3A) slope undefined; y-intercept = -3 B) slope = 0; y-intercept = -3C) slope undefined; no y-intercept D) slope = -3; y-intercept = 0


12) y = -3x

A) slope = 0; y-intercept = -3 B) slope = 3; y-intercept = 0

C) slope = -3; y-intercept = 0 D) slope = - 13; y-intercept = 0

8 Graph Lines Written in General Form Using Intercepts


Find the general form of the equation for the line with the given properties.

1) Slope = 35; y-intercept = 1

5

A) 3x + 5y = -1 B) 3x - 5y = -1 C) y = 35x + 1

5D) y = 3

5x - 1

5

2) Slope = - 37; containing the point (2, 3)

A) 3x + 7y = -27 B) 3x - 7y = 27 C) 7x + 3y = -27 D) 3x + 7y = 27

3) Slope = - 37; containing the point (0, 3)

A) 3x + 7y = -21 B) 3x - 7y = 21 C) 3x + 7y = 21 D) 7x + 3y = -21

4) Slope = 67; containing (0, 2)

A) -6x + 7y = -14 B) -6x - 7y = 14 C) -6x + 7y = 14 D) 7x - 6y = -14

Find the slope of the line and sketch its graph.5) 3x + 5y = 19

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


A) slope = - 53

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B) slope = 53

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C) slope = 35

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D) slope = - 35

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

6) 2x - 5y = -4

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


A) slope = - 52

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B) slope = - 25

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C) slope = 52

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D) slope = 25

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

Solve the problem.7) Find an equation in general form for the line graphed on a graphing utility.

A) y = -2x - 1 B) 2x + y = -1 C) x + 2y = -2 D) y = - 12x - 1


9 Find Equations of Parallel Lines


Find an equation for the line with the given properties.1) The solid line L contains the point (3, 4) and is parallel to the dotted line whose equation is y = 2x. Give the

equation for the line L in slope-intercept form.

x-5 5

y

5

-5

x-5 5

y

5

-5

A) y = 2x + b B) y = 2x + 1 C) y = 2x - 2 D) y - 4 = 2(x - 3)

2) Parallel to the line y = 4x; containing the point (7, 2)A) y = 4x + 26 B) y - 2 = 4x - 7 C) y = 4x - 26 D) y = 4x

3) Parallel to the line x - 4y = 4; containing the point (0, 0)

A) y = 14x + 4 B) y = - 1

4x C) y = - 3

4D) y = 1

4x

4) Parallel to the line -5x - y = 6; containing the point (0, 0)

A) y = -5x B) y = 15x C) y = 1

5x + 6 D) y = - 1

5x

5) Parallel to the line y = 6; containing the point (5, 9)A) y = 9 B) y = 5 C) y = 6 D) y = -9

6) Parallel to the line x = -5; containing the point (8, 2)A) x = 2 B) y = -5 C) x = 8 D) y = 2

7) Parallel to the line 9x + 2y = 44; containing the point (6, 1)A) 6x + 2y = 44 B) 9x - 2y = 56 C) 2x + 9y = 1 D) 9x + 2y = 56

8) Parallel to the line -4x + 5y = -3; x-intercept = 2A) 5x + 4y = 8 B) 5x + 4y = 10 C) -4x + 5y = -8 D) -4x + 5y = 10


10 Find Equations of Perpendicular Lines


Find an equation for the line with the given properties.1) The solid line L contains the point (1, 4) and is perpendicular to the dotted line whose equation is y = 2x. Give

the equation of line L in slope-intercept form.

x-5 5

y

5

-5

x-5 5

y

5

-5

A) y - 4 = 2(x - 1) B) y = 12x + 9

2C) y - 4 = - 1

2(x - 1) D) y = - 1

2x + 9

2

2) Perpendicular to the line y = -2x - 3; containing the point (3, 2)

A) y = 12x + 1

2B) y = -2x + 1

2C) y = 2x + 1

2D) y = - 1

2x + 1

2

3) Perpendicular to the line y = 17x + 8; containing the point (5, -2)

A) y = 7x - 33 B) y = - 7x + 33 C) y = - 17x - 33

7D) y = - 7x - 33

4) Perpendicular to the line -4x - y = 6; containing the point (0, - 32)

A) y = 14x + 6 B) y = 1

4x - 3

2C) y = - 1

4x - 3

2D) y = - 5

4

5) Perpendicular to the line x - 7y = 8; containing the point (4, 4)

A) y = 7x - 32 B) y = - 17x - 32

7C) y = - 7x + 32 D) y = - 7x - 32

6) Perpendicular to the line y = 4; containing the point (8, 3)A) y = 3 B) x = 8 C) x = 3 D) y = 8

7) Perpendicular to the line x = -2; containing the point (8, 9)A) y = 8 B) y = 9 C) x = 8 D) x = 9

8) Perpendicular to the line 3x - 4y = 35; containing the point (9, 2)A) 9x + 4y = 35 B) 4x + 3y = 42 C) 3x + 4 = 3 D) 4x - 3y = 42

9) Perpendicular to the line -4x - 7y = 20; containing the point (9, -15)A) -4x + 7y = -123 B) -7x - 4y = -123 C) -7x - 4y = 20 D) -7x + 4y = -123


10) Perpendicular to the line 3x - 2y = 3; y-intercept = 5A) 3x - 2y = -10 B) 3x - 2y = 15 C) -2x - 3y = -10 D) -2x - 3y = -15

Decide whether the pair of lines is parallel, perpendicular, or neither.11) 3x - 8y = 17

32x + 12y = 15A) parallel B) perpendicular C) neither

12) 3x - 6y = 218x + 9y = 16

A) parallel B) perpendicular C) neither

13) 6x + 2y = 812x + 4y = 17

A) parallel B) perpendicular C) neither

2.4 Circles

1 Write the Standard Form of the Equation of a Circle


Write the standard form of the equation of the circle.1)

x

y

(4, 4) (8, 4)

x

y

(4, 4) (8, 4)

A) (x + 6)2 + (y + 4)2 = 2 B) (x - 6)2 + (y - 4)2 = 4C) (x - 6)2 + (y - 4)2 = 2 D) (x + 6)2 + (y + 4)2 = 4


2)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) (x + 2)2 + (y + 3)2 = 9 B) (x - 2)2 + (y - 3)2 = 9C) (x - 3)2 + (y - 2)2 = 9 D) (x + 3)2 + (y + 2)2 = 9

Write the standard form of the equation of the circle with radius r and center (h, k).3) r = 3; (h, k) = (0, 0)

A) x2 + y2 = 3 B) x2 + y2 = 9C) (x - 3)2 + (y - 3)2 = 3 D) (x - 3)2 + (y - 3)2 = 9

4) r = 5; (h, k) = (7, -8)A) (x - 7)2 + (y + 8)2 = 25 B) (x + 7)2 + (y - 8)2 = 25C) (x + 7)2 + (y - 8)2 = 5 D) (x - 7)2 + (y + 8)2 = 5

5) r = 6; (h, k) = (-3, 0)A) (x + 3)2 + y2 = 36 B) (x - 3)2 + y2 = 36 C) x2 + (y - 3)2 = 6 D) x2 + (y + 3)2 = 6

6) r = 11; (h, k) = (0, 6)A) x2 + (y + 6)2 = 11 B) (x + 6)2 + y2 = 121 C) (x - 6)2 + y2 = 121 D) x2 + (y - 6)2 = 121

7) r = 10; (h, k) = (9, 1)A) (x + 1)2 + (y + 9)2 = 100 B) (x - 9)2 + (y - 1)2 = 10C) (x - 1)2 + (y - 9)2 = 100 D) (x + 9)2 + (y + 1)2 = 10

8) r = 14; (h, k) = (0, -9)A) x2 + (y - 9)2 = 14 B) (x - 9)2 + y2 = 196 C) (x + 9)2 + y2 = 196 D) x2 + (y + 9)2 = 14

Solve the problem.9) Find the equation of a circle in standard form where C(6, -2) and D(-4, 4) are endpoints of a diameter.

A) (x - 1)2 + (y - 1)2 = 34 B) (x + 1)2 + (y + 1)2 = 34C) (x + 1)2 + (y + 1)2 = 136 D) (x - 1)2 + (y - 1)2 = 136

10) Find the equation of a circle in standard form with center at the point (-3, 2) and tangent to the line y = 4.A) (x + 3)2 + (y - 2)2 = 16 B) (x - 3)2 + (y + 2)2 = 16C) (x + 3)2 + (y - 2)2 = 4 D) (x - 3)2 + (y + 2)2 = 4


11) Find the equation of a circle in standard form that is tangent to the line x = -3 at (-3, 5) and also tangent to theline x = 9.

A) (x - 3)2 + (y + 5)2 = 36 B) (x + 3)2 + (y - 5)2 = 36C) (x + 3)2 + (y + 5)2 = 36 D) (x - 3)2 + (y - 5)2 = 36

Find the center (h, k) and radius r of the circle with the given equation.12) x2 + y2 = 4

A) (h, k) = (0, 0); r = 4 B) (h, k) = (2, 2); r = 2C) (h, k) = (2, 2); r = 4 D) (h, k) = (0, 0); r = 2

13) (x - 3)2 + (y + 4)2 = 64A) (h, k) = (3, -4); r = 8 B) (h, k) = (-4, 3); r = 64C) (h, k) = (3, -4); r = 64 D) (h, k) = (-4, 3); r = 8

14) (x - 2)2 + y2 = 9A) (h, k) = (0, 2); r = 9 B) (h, k) = (2, 0); r = 3C) (h, k) = (0, 2); r = 3 D) (h, k) = (2, 0); r = 9

15) x2 + (y - 9)2 = 81A) (h, k) = (0, 9); r = 81 B) (h, k) = (0, 9); r = 9C) (h, k) = (9, 0); r = 9 D) (h, k) = (9, 0); r = 81

16) 5(x + 6)2 + 5(y + 1)2 = 45A) (h, k) = (6, 1); r = 15 B) (h, k) = (-6, -1); r = 3C) (h, k) = (6, 1); r = 3 D) (h, k) = (-6, -1); r = 15

Solve the problem.17) Find the standard form of the equation of the circle. Assume that the center has integer coordinates and the

radius is an integer.

A) (x - 1)2 + (y + 2)2 = 9 B) x2 + y2 + 2x - 4y - 4 = 0C) (x + 1)2 + (y - 2)2 = 9 D) x2 + y2 - 2x + 4y - 4 = 0


2 Graph a Circle


Graph the circle with radius r and center (h, k).1) r = 3; (h, k) = (0, 0)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


2) r = 3; (h, k) = (0, 4)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


3) r = 4; (h, k) = (3, 0)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


4) r = 2; (h, k) = (5, 5)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


Graph the equation.5) x2 + y2 = 16

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


6) (x + 1)2 + (y - 5)2 = 16

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


7) x2 + (y - 1)2 = 36

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


8) (x - 6)2 + y2 = 9

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D)

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


3 Work with the General Form of the Equation of a Circle


Find the center (h, k) and radius r of the circle. Graph the circle.1) x2 + y2 - 10x - 10y + 25 = 0

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) (h, k) = (-5, -5); r = 5

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B) (h, k) = (5, -5); r = 5

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C) (h, k) = (-5, 5); r = 5

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D) (h, k) = (5, 5); r = 5

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10


2) x2 + y2 + 6x + 8y + 9 = 0

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

A) (h, k) = (3, 4); r = 4

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

B) (h, k) = (-3, -4); r = 4

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

C) (h, k) = (-3, 4); r = 4

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

D) (h, k) = (3, -4); r = 4

x-10 -5 5 10

y10

5

-5

-10

x-10 -5 5 10

y10

5

-5

-10

Find the center (h, k) and radius r of the circle with the given equation.3) x2 - 16x + 64 + (y + 3)2 = 49

A) (h, k) = (8, -3); r = 7 B) (h, k) = (-8, 3); r = 49C) (h, k) = (-3, 8); r = 7 D) (h, k) = (3, -8); r = 49

4) x2 + 12x + 36 + y2 - 16y + 64 = 4A) (h, k) = (6, -8); r = 4 B) (h, k) = (-8, 6); r = 4C) (h, k) = (8, -6); r = 2 D) (h, k) = (-6, 8); r = 2

5) x2 + y2 - 16x - 2y + 65 = 49A) (h, k) = (1, 8); r = 7 B) (h, k) = (-1, -8); r = 49C) (h, k) = (-8, -1); r = 49 D) (h, k) = (8, 1); r = 7


6) x2 + y2 + 12x + 14y = -60A) (h, k) = (-7, -6); r = 5 B) (h, k) = (7, 6); r = 25C) (h, k) = (6, 7); r = 25 D) (h, k) = (-6, -7); r = 5

7) 4x2 + 4y2 - 12x + 16y - 5 = 0

A) (h, k) = ( 32, -2); r = 3 5

2B) (h, k) = (- 3

2, 2); r= 3 5

2

C) (h, k) = ( 32, -2); r = 30

2D) (h, k) = (- 3

2, 2); r = 30

2

Find the general form of the equation of the the circle.8) Center at the point (-4, -3); containing the point (-3, 3)

A) x2 + y2 - 6x + 6y - 12 = 0 B) x2 + y2 + 6x - 6y - 17 = 0C) x2 + y2 + 6x + 8y - 17 = 0 D) x2 + y2 + 8x + 6y - 12 = 0

9) Center at the point (2, -3); containing the point (5, -3)A) x2 + y2 + 4x - 6y + 4 = 0 B) x2 + y2 + 4x - 6y + 22 = 0C) x2 + y2 - 4x + 6y + 4 = 0 D) x2 + y2 - 4x + 6y + 22 = 0

10) Center at the point (-3, -1); tangent to y-axisA) x2 + y2 + 6x + 2y + 1 = 0 B) x2 + y2 + 6x + 2y + 9 = 0C) x2 + y2 - 6x - 2y + 9 = 0 D) x2 + y2 + 6x + 2y + 11 = 0

Solve the problem.11) If a circle of radius 5 is made to roll along the x-axis, what is the equation for the path of the center of the

circle?A) y = 10 B) x = 5 C) y = 5 D) y = 0

12) Earth is represented on a map of the solar system so that its surface is a circle with the equationx2 + y2 + 8x + 2y - 4079 = 0. A weather satellite circles 0.6 units above the Earth with the center of its circularorbit at the center of the Earth. Find the general form of the equation for the orbit of the satellite on this map.

A) x2 + y2 - 8x - 2y - 4156.16 = 0 B) x2 + y2 + 8x + 2y - 46.64 = 0C) x2 + y2 + 8x + 2y + 16.64 = 0 D) x2 + y2 + 8x + 2y - 4156.16 = 0

13) Find an equation of the line containing the centers of the two circlesx2 + y2 + 2x + 2y + 1 = 0 andx2 + y2 + 6x + 8y + 21 = 0

A) 3x - 2y + 1 = 0 B) -3x - 2y + 1 = 0 C) 3x + 2y + 1 = 0 D) -5x + 4y + 1 = 0

14) A wildlife researcher is monitoring a black bear that has a radio telemetry collar with a transmitting range of 24miles. The researcher is in a research station with her receiver and tracking the bearʹs movements. If we put theorigin of a coordinate system at the research station, what is the equation of all possible locations of the bearwhere the transmitter would be at its maximum range?

A) x2 + y2 = 24 B) x2 + y2 = 576 C) x2 - y2 = 24 D) x2 + y2 = 48

15) If a satellite is placed in a circular orbit of 290 kilometers above the Earth, what is the equation of the path of thesatellite if the origin is placed at the center of the Earth (the diameter of the Earth is approximately 12,740kilometers)?

A) x2 + y2 = 84,100 B) x2 + y2 = 169,780,900C) x2 + y2 = 40,576,900 D) x2 + y2 = 44,355,600


16) A power outage affected all homes and businesses within a 11 mi radius of the power station. If the powerstation is located 13 mi north of the center of town, find an equation of the circle consisting of the furthestpoints from the station affected by the power outage.

A) x2 + (y - 13)2 = 121 B) x2 + (y - 13)2 = 11C) x2 + (y + 13)2 = 121 D) x2 + y2 = 121

17) A power outage affected all homes and businesses within a 2 mi radius of the power station. If the powerstation is located 3 mi west and 3 mi north of the center of town, find an equation of the circle consisting of thefurthest points from the station affected by the power outage.

A) (x - 3)2 + (y + 3)2 = 4 B) (x - 3)2 + (y - 3)2 = 4C) (x + 3)2 + (y + 3)2 = 4 D) (x + 3)2 + (y - 3)2 = 4

18) A Ferris wheel has a diameter of 400 feet and the bottom of the Ferris wheel is 10 feet above the ground. Findthe equation of the wheel if the origin is placed on the ground directly below the center of the wheel, asillustrated.

400 ft.

10 ft.

A) x2 + (y - 210)2 = 40,000 B) x2 + (y - 200)2 = 160,000C) x2 + y2 = 40,000 D) x2 + (y - 200)2 = 40,000

2.5 Variation

1 Construct a Model Using Direct Variation


Write a general formula to describe the variation.1) v varies directly with t; v = 16 when t = 20

A) v = 54t B) v = 20

16tC) v = 16

20tD) v = 4

5t

2) A varies directly with t2; A = 245 when t = 7

A) A = 35t2

B) A = 35t2 C) A = 5t2

D) A = 5t2

3) z varies directly with the sum of the squares of x and y; z = 5 when x = 3 and y = 4

A) z = 15(x2 + y2) B) z2 = x2 + y2 C) z = 1

10(x2 + y2) D) z = 1

25(x2 + y2)


If y varies directly as x, write a general formula to describe the variation.4) y = 5 when x = 45

A) y = 9x B) y = 15x C) y = 1

9x D) y = x + 40

5) y = 15 when x = 27

A) y = 3x B) y = x - 12 C) y = 95x D) y = 5

9x

6) y = 3 when x = 16

A) y = 118

x B) y = 18x C) y = 13x D) y = x + 17

6

7) y = 1.5 when x = 0.3A) y = 5x B) y = 0.2x C) y = x + 1.2 D) y = 0.3x

8) y = 0.2 when x = 1.6A) y = x - 1.4 B) y = 0.2x C) y = 0.125x D) y = 8x

Write a general formula to describe the variation.9) The volume V of a right circular cone varies directly with the square of its base radius r and its height h. The

constant of proportionality is 13π.

A) V = 13πrh B) V = 1

3r2h C) V = 1

3πr2h2 D) V = 1

3πr2h

10) The surface area S of a right circular cone varies directly as the radius r times the square root of the sum of thesquares of the base radius r and the height h. The constant of proportionality is π.

A) S = πr r2h2 B) S = πr r2h C) S = πr r2 + h2 D) S = π r2 + h2

Solve the problem.11) In simplified form, the period of vibration P for a pendulum varies directly as the square root of its length L. If

P is 1 sec. when L is 4 in., what is the period when the length is 100 in.?A) 200 sec B) 20 sec C) 50 sec D) 5 sec

12) The amount of water used to take a shower is directly proportional to the amount of time that the shower is inuse. A shower lasting 17 minutes requires 8.5 gallons of water. Find the amount of water used in a showerlasting 5 minutes.

A) 10 gal B) 28.9 gal C) 1.7 gal D) 2.5 gal

13) If the resistance in an electrical circuit is held constant, the amount of current flowing through the circuit isdirectly proportional to the amount of voltage applied to the circuit. When 9 volts are applied to a circuit,180 milliamperes (mA) of current flow through the circuit. Find the new current if the voltage is increased to15 volts.

A) 135 mA B) 320 mA C) 285 mA D) 300 mA

14) The amount of gas that a helicopter uses is directly proportional to the number of hours spent flying. Thehelicopter flies for 2 hours and uses 18 gallons of fuel. Find the number of gallons of fuel that the helicopteruses to fly for 6 hours.

A) 54 gal B) 60 gal C) 63 gal D) 12 gal


15) The distance that an object falls when it is dropped is directly proportional to the square of the amount of timesince it was dropped. An object falls 128 feet in 2 seconds. Find the distance the object falls in 4 seconds.

A) 8 ft B) 512 ft C) 128 ft D) 256 ft

2 Construct a Model Using Inverse Variation


Write a general formula to describe the variation.1) A varies inversely with x2; A = 7 when x = 2

A) A = 14x2

B) A = 28x2

C) A = 74x2 D) A = 14x2

Write an equation that expresses the relationship. Use k as the constant of variation.2) c varies inversely as t.

A) c = kt B) kc = t C) c = kt

D) c = tk

3) r varies inversely as the square of t.

A) r = t2k

B) r = tk

C) r = kt2

D) r = kt

If y varies inversely as x, write a general formula to describe the variation.4) y = 9 when x = 7

A) y = 97x B) y = 1

63xC) y = 63

xD) y = x

63

5) y = 55 when x = 8

A) y = 558x B) y = 1

440xC) y = x

440D) y = 440

x

6) y = 36 when x = 19

A) y = 14x

B) y = x4

C) y = 324x D) y = 4x

7) y = 12 when x = 18

A) y = 136

x B) y = 19x

C) y = x9

D) y = 9x

8) y = 0.5 when x = 0.2

A) y = 10x B) y = 10x

C) y = 0.1x

D) y = 2.5x

Solve the problem.9) x varies inversely as v, and x = 12 when v = 2. Find x when v = 4.

A) x = 4 B) x = 12 C) x = 6 D) x = 2

10) x varies inversely as y2, and x = 2 when y = 30. Find x when y = 5.A) x = 50 B) x = 72 C) x = 6 D) x = 24


11) When the temperature stays the same, the volume of a gas is inversely proportional to the pressure of the gas.If a balloon is filled with 87 cubic inches of a gas at a pressure of 14 pounds per square inch, find the newpressure of the gas if the volume is decreased to 29 cubic inches.

A) 42 psi B) 39 psi C) 2914 psi D) 28 psi

12) The amount of time it takes a swimmer to swim a race is inversely proportional to the average speed of theswimmer. A swimmer finishes a race in 300 seconds with an average speed of 4 feet per second. Find theaverage speed of the swimmer if it takes 400 seconds to finish the race.

A) 4 ft/sec B) 3 ft/sec C) 5 ft/sec D) 2 ft/sec

13) If the force acting on an object stays the same, then the acceleration of the object is inversely proportional to itsmass. If an object with a mass of 50 kilograms accelerates at a rate of 5 meters per second per second (m/sec2)by a force, find the rate of acceleration of an object with a mass of 5 kilograms that is pulled by the same force.

A) 45 m/sec2 B) 50 m/sec2 C) 40 m/sec2 D) 12 m/sec2

14) If the voltage, V, in an electric circuit is held constant, the current, I, is inversely proportional to the resistance,R. If the current is 360 milliamperes (mA) when the resistance is 5 ohms, find the current when the resistance is30 ohms.

A) 2154 mA B) 60 mA C) 300 mA D) 2160 mA

15) While traveling at a constant speed in a car, the centrifugal acceleration passengers feel while the car is turningis inversely proportional to the radius of the turn. If the passengers feel an acceleration of 12 feet per second persecond (ft/sec2) when the radius of the turn is 40 feet, find the acceleration the passengers feel when the radiusof the turn is 160 feet.

A) 3 ft/sec2 B) 6 ft/sec2 C) 4 ft/sec2 D) 5 ft/sec2

3 Construct a Model Using Joint Variation or Combined Variation


Write a general formula to describe the variation.1) The square of G varies directly with the cube of x and inversely with the square of y; G = 2 when x = 3 and

y = 3

A) G2 = 2 x3

y2B) G2 = 12 y

3

x2C) G2 = 4

3 x3

y2D) G2 = 4

243 (x3 + y2)

2) R varies directly with g and inversely with the square of h; R = 3 when g = 3 and h = 5.

A) R = 25 gh2

B) R = 5 h2g

C) R = 25gh2 D) R = 5 gh2

3) z varies jointly as the cube root of x and the cube of y; z = 1 when x = 8 and y = 5.

A) z = 250 3xy3 B) z = 1

250 3xy3 C) z = 2

125 3x

y3D) z = 125

2

3x

y3

4) The centrifugal force F of an object speeding around a circular course varies directly as the product of theobjectʹs mass m and the square of itʹs velocity v and inversely as the radius of the turn r.

A) F = kmv2r

B) F = km2vr

C) F = kmrv2

D) F = kmvr


5) The safety load λ of a beam with a rectangular cross section that is supported at each end varies directly as theproduct of the width W and the square of the depth D and inversely as the length L of the beam between thesupports.

A) λ = k(W + D2)

LB) λ = kL

WD2C) λ = kWD2

LD) λ = kWD

L

6) The illumination I produced on a surface by a source of light varies directly as the candlepower c of the sourceand inversely as the square of the distance d between the source and the surface.

A) I = kcd2

B) I = kd2c

C) I = kcd2 D) I = kc2

d2

Solve the problem.7) The volume V of a given mass of gas varies directly as the temperature T and inversely as the pressure P. A

measuring device is calibrated to give V = 88.4 in3 when T = 170° and P = 25 lb/in2. What is the volume on thisdevice when the temperature is 420° and the pressure is 20 lb/in2?

A) V = 313 in3 B) V = 273 in3 C) V = 233 in3 D) V = 21 in3

8) The time in hours it takes a satellite to complete an orbit around the earth varies directly as the radius of theorbit (from the center of the earth) and inversely as the orbital velocity. If a satellite completes an orbit810 miles above the earth in 16 hours at a velocity of 38,000 mph, how long would it take a satellite to completean orbit if it is at 1800 miles above the earth at a velocity of 26,000 mph? (Use 3960 miles as the radius of theearth.)

A) 8.82 hr B) 28.24 hr C) 282.38 hr D) 51.97 hr

9) The pressure of a gas varies jointly as the amount of the gas (measured in moles) and the temperature andinversely as the volume of the gas. If the pressure is 810 kiloPascals (kPa) when the number of moles is 5, thetemperature is 270° Kelvin, and the volume is 400 cc, find the pressure when the number of moles is 2, thetemperature is 340° K, and the volume is 120 cc.

A) 680 kPa B) 1440 kPa C) 1360 kPa D) 640 kPa

10) Body-mass index, or BMI, takes both weight and height into account when assessing whether an individual isunderweight or overweight. BMI varies directly as oneʹs weight, in pounds, and inversely as the square of oneʹsheight, in inches. In adults, normal values for the BMI are between 20 and 25. A person who weighs 182pounds and is 67 inches tall has a BMI of 28.5. What is the BMI, to the nearest tenth, for a person who weighs137 pounds and who is 64 inches tall?

A) 23.1 B) 23.5 C) 23.9 D) 22.7

11) The amount of paint needed to cover the walls of a room varies jointly as the perimeter of the room and theheight of the wall. If a room with a perimeter of 70 feet and 10-foot walls requires 7 quarts of paint, find theamount of paint needed to cover the walls of a room with a perimeter of 65 feet and 6-foot walls.

A) 3.9 qt B) 7.8 qt C) 390 qt D) 39 qt

12) The power that a resistor must dissipate is jointly proportional to the square of the current flowing through theresistor and the resistance of the resistor. If a resistor needs to dissipate 32 watts of power when 4 amperes ofcurrent is flowing through the resistor whose resistance is 2 ohms, find the power that a resistor needs todissipate when 2 amperes of current are flowing through a resistor whose resistance is 8 ohms.

A) 16 watts B) 64 watts C) 32 watts D) 128 watts


13) While traveling in a car, the centrifugal force a passenger experiences as the car drives in a circle varies jointlyas the mass of the passenger and the square of the speed of the car. If a passenger experiences a force of 57.6newtons (N) when the car is moving at a speed of 40 kilometers per hour and the passenger has a mass of 40kilograms, find the force a passenger experiences when the car is moving at 80 kilometers per hour and thepassenger has a mass of 50 kilograms.

A) 256 N B) 288 N C) 320 N D) 384 N

14) The amount of simple interest earned on an investment over a fixed amount of time is jointly proportional tothe principle invested and the interest rate. A principle investment of $3000.00 with an interest rate of 3%earned $360.00 in simple interest. Find the amount of simple interest earned if the principle is $4200.00 and theinterest rate is 7%.

A) $117,600.00 B) $1176.00 C) $504.00 D) $840.00

15) The voltage across a resistor is jointly proportional to the resistance of the resistor and the current flowingthrough the resistor. If the voltage across a resistor is 45 volts (V) for a resistor whose resistance is 5 ohms andwhen the current flowing through the resistor is 9 amperes, find the voltage across a resistor whose resistanceis 7 ohms and when the current flowing through the resistor is 3 amperes.

A) 15 V B) 27 V C) 63 V D) 21 V


Ch. 2 GraphsAnswer Key

2.1 The Distance and Midpoint Formulas1 Rectangular Coordinates

1) A2) B3) C4) D5) C6) C7) C8) B9) D10) B11) D12) C13) C14) C15) B16) B17) D18) A

2 Use the Distance Formula1) D2) D3) A4) C5) C6) A7) D8) D9) C10) D11) B12) C13) A14) A15) B16) B17) A18) A19) B20) D21) B22) B23) B

3 Use the Midpoint Formula1) B2) B3) A4) A5) D6) C


7) C8) A9) A10) A11) C

2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry1 Graph Equations by Plotting Points

1) A2) B3) C4) D5) B6) C7) D8) C9) D10) A

2 Find Intercepts from a Graph1) D2) D3) B4) D5) C6) A7) A8) A

3 Find Intercepts from an Equation1) D2) A3) A4) D5) C6) A7) B8) D9) D10) D11) B12) B13) C

4 Test an Equation for Symmetry with Respect to the x-Axis, the y-Axis, and the Origin1) D2) C3) B4) B5) B6) A7) C8) D9) A10) C11) B12) C13) C


14) E15) C16) B17) C18) D19) D20) E21) C22) C23) B24) E25) C26) E27) A28) C

5 Know How to Graph Key Equations1) C2) B3) C4) C

2.3 Lines1 Calculate and Interpret the Slope of a Line

1) A2) C3) A4) D5) B6) C7) D8) A9) D10) B

2 Graph Lines Given a Point and the Slope1) B2) C3) D4) B5) C6) B7) B8) C9) C

3 Find the Equation of a Vertical Line1) B2) A3) C4) C

4 Use the Point-Slope Form of a Line; Identify Horizontal Lines1) C2) D3) C4) A5) D


5 Find the Equation of a Line Given Two Points1) B2) C3) D4) C5) C6) D7) D8) B9) C10) C11) B12) C13) A14) D15) C16) C

6 Write the Equation of a Line in Slope-Intercept Form1) C2) D3) C4) A5) D6) C7) D8) A9) C10) B11) D12) A13) D

7 Identify the Slope and y-Intercept of a Line from Its Equation1) D2) C3) C4) A5) D6) A7) D8) D9) D10) D11) C12) C

8 Graph Lines Written in General Form Using Intercepts1) B2) D3) C4) C5) D6) D7) C

9 Find Equations of Parallel Lines1) C


2) C3) D4) A5) A6) C7) D8) C

10 Find Equations of Perpendicular Lines1) D2) A3) B4) B5) C6) B7) B8) B9) D10) D11) B12) B13) A

2.4 Circles1 Write the Standard Form of the Equation of a Circle

1) B2) C3) B4) A5) A6) D7) B8) D9) A10) C11) D12) D13) A14) B15) B16) B17) C

2 Graph a Circle1) B2) A3) B4) C5) B6) B7) B8) B

3 Work with the General Form of the Equation of a Circle1) D2) B3) A4) D


5) D6) D7) C8) D9) C10) B11) C12) D13) A14) B15) D16) A17) D18) A

2.5 Variation1 Construct a Model Using Direct Variation

1) D2) D3) A4) C5) D6) B7) A8) C9) D10) C11) D12) D13) D14) A15) B

2 Construct a Model Using Inverse Variation1) B2) C3) C4) C5) D6) D7) D8) C9) C10) B11) A12) B13) B14) B15) A

3 Construct a Model Using Joint Variation or Combined Variation1) C2) A3) B4) A5) C6) A


7) B8) B9) C10) B11) A12) C13) B14) B15) D


ch. 2 graphs...f - ; - f + ; c

Documents