Intermediate Algebra (B) Unit 5 – Selected Answers · 2015-01-08 · Intermediate Algebra (B)...

Intermediate Algebra (B) Unit 5 – Selected Answers

Answers Intermediate Algebra (B) ~ page 1 ~

SECTION 5.1A

1) “Find the zeros of the function”

means to find the x-value(s)

that make the function f(x) = 0.

(answers will vary)

2) The x-intercepts (answers will

vary)

3) # & type: 1 real root (rational)

zeros: x = 3

4) # & type: no real roots

zeros: none

5) # & type: 2 real roots (rational)

zeros: x = 0 or x = –2

6) solutions: x = 3 or x = –5

check: f(3) = (3)2 + 2(3) – 15 = 0

f(–5) = (–5)2 + 2(–5) – 15 = 0

both solutions create a zero

7) solutions: x = 0 or x = 4

check: f(0) = –2(0)2 + 8(0) = 0

f(4) = –2(4)2 + 8(4) = 0

both solutions create a zero

8)

solutions: x = –2 or x = 6

check:

f(–2) = –(–2)2 + 4(–2) + 12 = 0

f(6) = –(6)2 + 4(6) + 12 = 0

9)

solutions: x = –5 or x = 2

check:

f(–5) = (–5)2 + 3(–5) – 10 = 0

f(2) = (2)2 + 3(2) – 10 = 0

10) solutions: x ≈ –1.32 or x ≈ 8.32

11) solutions: x ≈ –1.67or x = –1.50

12) solutions: x ≈ –0.94 or x ≈ 2.34

13) solutions: x ≈ –3.12 or x ≈ 5.12

0.5(−3.12) − (−3.12) − 8 ≈ 0

0.5(5.12) − (5.12) − 8 ≈ 0

14) solutions: x ≈ –5.16 or x ≈ 1.16

(−5.16) + 4(−5.16) ≈ 6

(1.16) + 4(1.16) ≈ 6

15) solutions: x ≈ –1.16 or x ≈ 2.16

2(−1.16) − 2(−1.16) − 5 ≈ 0

2(2.16) − 2(2.16) − 5 ≈ 0

16)

solutions: x = –5 or x = –1

17)


18)


19)

solutions: x = 0 or x = –8



SECTION 5.1A (continued)

20)


21)


22) a) (16) f(x) = x2 + 6x + 5

x = –5, x = –1

(17) f(x) = 5x2 + 30x + 25

x = –5, x = –1

(18) f(x) = 3x2 + 18x + 15

x = –5, x = –1

b) (19) f(x) = x2 + 8x

x = 0, x = –8

(20) f(x) = (½)x2 + 4x

x = 0, x = –8

(21) f(x) = 2x2 + 16x

x = 0, x = –8

c) If one function is a multiple

of another function, then

they will have the same

solutions. (answers will

vary)

23) a)

b) 5 feet; the point (5, 5) is the

vertex of the graph of the

equation.

c) 10 feet; this is the x-intercept

of the graph of the equation

(that is greater than zero).

(answers will vary)

SECTION 5.1B

1) a) Peter is correct. The

x-intercept gives the time

(t) when y = 0, which is

where the ball has a height

of 0 on the ground.

(explanations will vary)

b) Pablo’s ball hit the ground

first. It took 2.5 seconds

for it to hit the ground.

c) Peter’s ball hit the ground

second. It took 3.125

seconds for it to hit the

ground.

2) a) h(t) = –16t2 + 35t + 5.5

b) 24.5 feet

c) After 0.14 seconds and

then again after 2.05

seconds.

d) After 1.09 seconds.

2) e) 24.64 feet is the maximum

that the football ever gets

into the air.

f) 2.33 seconds is when the

ball hits the ground

(the x-intercept).

3) Vertex: (1.56, 41.06). The

vertex reveals how long it

takes (1.56 seconds) for the

ball to reach the maximum

height (41.06 ft).


4) t = 11 days, which is the

x-coordinate of the vertex.

5) a) 1.5 sections is the

maximum speed (the

x-coordinate of the vertex

of the graph)

b) 89 km/h (the y-coordinate

of the vertex of the graph)

5) c) 6.22 seconds (the

x-coordinate of the

x-intercept of the graph.

6) a) 50 is half the perimeter, so

one length plus one width

equals 50 feet of fencing.


b) A(L) = L(50 – L)

A(L) = –L2 + 50L

c) (L, A(L)) = (25, 625)

when the length is 25, the

width is 50 – L = 25. So,

Area = L × W

25 ft × 25 ft

625 ft2

Area of the tomato patch is

625 square feet.

d) The tomato patch is a

square shape (25ft × 25ft).



SECTION 5.1B (continued)

7) a) According to #6d, Sharon

should make a square with

her 40 feet of fencing. So,

40 ÷ 4 = 10 feet on each

side.

b) A(L) = L(20 – L)

A(L) = –L2 + 20L

calc the max, shows a

vertex at (10, 100)

7 c) (L, A(L)) = (10, 100)

when the length is 10, the

width is 20 – L = 10. So,

Area = L × W

10 ft × 10 ft = 100 ft2

The area of the fenced play

area is 100 square feet.

8) Yes! The maximum height of

the orange is 19 feet (which is

the y-coordinate of the vertex

of the graph). Jim can grab

the orange either on the way

up for the orange or on its way

down (explanations will vary).

9) At approximately 15 miles per

hour will the car be able to

stop for a sign 30 feet away.

10) a) 141.56 feet

b) 5.79 seconds

SECTION 5.2A

1) 2(x + 3)

2) 3(y – 3)

3) 7(a + 4)

4) 12(3z – 1)

5) b(b + 1)

6) r(2 – r)

7) t(9t + 1)

8) n(4n – 5)

9) 4h(h + 3)

10) 9x(1 – 3x)

11) 2a(a + 2)

12) 4d(5d – 6)

13) (x + 7)(x + 6)

14) (x + 3)(x + 3) or (x + 3)2

15) (x + 8)(x + 4)

16) (x + 5)(x – 2)

17) (x – 5)(x – 5) or (x – 5)2

18) (x – 4)(x + 3)

19) (3x + 4)(x – 1)

20) (2x – 3)(x + 4)

21) (2x – 3)( 2x – 3) or (2x – 3)2

22) (6x – 1)(2x – 1)

23) 2(x + 1)(x – 5)

24) 3(x + 1)(x + 6)

25) (a – 3)(a + 3)

26) (x – 4)(x + 4)

27) (5x – 6)(5x + 6)

28) y = (x + 1)(x + 2)

29) y = (x + 7)(x – 7)

30) The x-intercepts have a

y-coordinate of zero.

Setting y = 0 and solving for

x gives the solution(s),

which are the x-intercept(s).

(answers will vary)

31) (4, 0) and (20, 0)

32) (1, 0) and (–10, 0)

33) a) an arrow should be drawn to

the location that the graph of

the height of the ball in

relationship to time touches

the x-axis.

33) b) y = 0 means the place where

the ball hits the ground

(where the height is zero).

SECTION 5.2B

1) x = 4 or x = –9

2) x = 2 (double root)

3) 3

4x = − or

5

2x =

4) x = 3 or x = –3

5) x = 2 or x = –2

6) x = 0 or x = 3

7) x = 0 or x = 4

8) x = 1 or x = –9

9) x = –4 or x = –3

10) x = 5 (double root)

11) x = –1 or x = 4

12) x = –1 or x = 5

13) x = 5 or x = 8

14) 4

3x = − or x = –2

15) 5

4x = − or

1

2x =

16) 1

5x = − or x = –2

17) 1

2x = − or x = 8



SECTION 5.2C

1) The legs of the triangle have

lengths of 3 units and 4 units.

2) The legs of the triangle have

lengths of 5 units and 12 units.

3) The value of x is 10 (side

lengths are 9 units and 12

units).

4) The value of x is 11 (side

lengths are 10 units and 12

units).

5) The two numbers that satisfy

the situation are 8 and 15.

6) The dimensions are 14 feet by

48 feet.

7) The two positive numbers are

5 and 3.

8) The two negative integers are

–4 and –6.

9) (x = 2) You can frame a

2ft × 2ft square picture.

SECTION 5.2D

1) a) 2 5

b) 4 3

c) 10 2

d) 6 5

e) 10 6

f) 42 2

2) a) 9

b) 6 11 2+

c) 2 10 7−

d) 5 5 2−

e) 3 5−

f) 11 8 3+

3) a)

3) b)

4) a) 5

b) 3 10

c) 3 2

d) 12 15

e) –80

f) 6 42−

5) a) 15 5 7+

b) 6 3 3− +

c) 2 8 3+

d) 2 5 3 10+

e) 6 3 2− −

f) 5 3 2 6− +

6) a) 19 8 3+

b) 22 12 2−

c) 47 26 3−

d) 76 24 2+

e) 132 48 7−

f) 42 7 11−

7) a) yes, this x-value is a

solution

b) no (x = 6 2+ ) is not a

solution to the equation.

c) yes, (x = 5 3 5− ) is a


8) a) 2a + 5

b) 2 5 2+

c) 7 5 6− −

d) 5 6 10− +

9) 2

?

?

?

?

?

?

?

?

3 57 3 572 3 6 0

4 4

9 6 57 57 3 572 3 6 0

16 4

66 6 57 9 3 572 6 0

16 4

33 3 57 9 3 572 6 0

8 4

33 3 57 9 3 576 0

4 4

33 9 3 57 3 576 0

4

246 0

4

6 6 0

0 0

+ +− − =

+ + +− − =

+ − −+ − =

+ − −+ − =

+ − −+ − =

− + −− =

− =

− =

=



SECTION 5.2E

1) x = –2 or x = 2 ( 2x = ± )

2) x = –4 or x = 4 ( 4x = ± )

3) x = –5 or x = 5 ( 5x = ± )

4) x = –5 or x = 7

5) x = –7 or x = 1

6) x = –1 or x = 5

7) It takes 2 seconds to reach the

ground.

8) A sign should be posted

stating the maximum speed is

28 km per hour.

9) The other leg of the triangle

Andy created is 9 feet long.

SECTION 5.2F

1) The error occurs when Omar

tried to add 3 to both sides

before he takes the square root

of both sides. The correct

solutions are x = 6 or x = 0

2) Yes, both of the values given

are solutions.


are solutions.


are solutions.

5) No, neither value given is a

solution.

6) Both answers are correct. The

balloon could have hit a 6 foot

tall student on the way up

after 0.15 seconds, or it could

have hit the student on its

descending path after 1.15

seconds.


7) 3 2x = ±

8) 1 4 2x = ±

9) 2x = ±

10) x = –2 or x = 4

11) 5 3x = ±

12) x = –2

13) x = –2 or x = 4

14) 2 2x = ±

15) 2.41t ≈ seconds

16) 49.74x ≈ miles

SECTION 5.2G

1) a) 1−

b) 1

c) i

d) 1

e) i

f) i−

g) 5i

h) 12i

i) 10 7i

j) 14 2i

k) 4 10i−

l) 50i

2) a) 9 4i+

b) 13 6i−

2) c) 2 2i− +

d) 11 3i+

e) 9 3i−

f) 12 19i−

3) a) 6−

b) 12

c) 24 2i

d) 60i−

e) 48i

f) 18 2

4) (all answers in #4 may vary)

a) (1) in method 1, Kasem

never used the definition on

1i = − to rewrite either

radical. Method 3 uses the

definition correctly.

(2) method 1 used the

product property of radicals

incorrectly; only if a and b

are both nonnegative with an

even index, is

a b a b⋅ = ⋅ . Method 2

uses the product property of

radicals correctly.

b) (1) both methods combine

4 9 36⋅ = instead of

simplifying each separately.

(2) both methods used the

fact that 36 6=



SECTION 5.2G (continued)

4) c) (1) from line 3 to line 4,

multiplication was done in a

different order by

commutative property for

multiplication.

(2) method 3 simplified the

radicals 4 and 9 ;

method 2 multiplied them

together.

d) (1) both used the definition

of 1i = − whenever there

was a 1− .

(2) both used the fact that

2 1i = −

5) a) 2 3−

b) 12 10−

c) 63 2

6) a) 5 15i−

b) 9 3i+

c) 6 60i−

d) 12 6i−

e) 10 40i+

f) 2 5 2 10i+

7) a) 16 30i− +

b) 36 48i−

c) 7 30 2i−

8) a) 68 24 2i− +

b) 36 48 7i− −

c) 24 7 11i− −

9) a) Yes

−2(5�) + 6 = 56 56 = 56

b) Yes (4 + 3�) − 8(4 + 3�) + 30 = 5

5 = 5

9) c) Yes

( ) ( )

( ) ( )( ) ( ) ( )

2

1 3 5 2 1 3 5 48 2

1 3 5 1 3 5 2 6 5 48 2

1 6 5 9 1 5 2 6 5 48 2

2 2

i i

i i i

i i

− − − + =

− − − + + =

− + − − + + =

=

10) a) 3 4i+

b) 3 5 2i+

c) 5 7 6i− −

d) 6 3i−

11)

( )

( )

2?

2 ?

?

?

?

?

3 32 6 8 3

2 2

9 62 3 3 8 3

4

9 6 12 9 3 8 3

4

8 62 1 3 3

4

4 32 1 3 3

2

4 3 1 3 3

3 3

i i

i ii

ii

ii

ii

i i

+ + − + =

+ +− + + =

+ + −− − + =

+ − − =

+ − − =

+ − − =

=

SECTION 5.2H

1) 3i

2) 35 2i

3) 2 3i

4) 30i

5) 33 2i

6) 11i

7) –1

8) –9

9) –25

10) –1

11) 16 30i i− −

12) 14 48i− +

13) 40 28i− +

14) Yes

−2(3�) + 3 = 21 21 = 21

15) Yes

�(5 + 4�) − 5� − 1 = −17 −17 = −17

16) Yes

(2�) + 11 = 7 7 = 7

17) Yes

�(−2 + 5�) + 2� = −25 −25 = −25

18) � = ±4√3

19) � = 1 ± 2�√6

20) � = ±3

21) � = −1 ± 3� 22) � ≈ 1.86 seconds. The balloon lands on the ground after approximately 1.86 seconds. (–0.96 does not work when verified)

(#23-26, verification of solutions should be shown by student.)

23) � = ±5√5

24) � = −7

25) � = 2 ± 3� 26) � = ±�√6

27) 7.071s ≈ inches; The length of one side of the square is approximately 7.071 inches.

28) � = ±2�. With these solutions being imaginary, it reinforces the reason why the graph does not cross the x-axis. There are no real solutions, there are two complex solutions.



SECTION 5.2I

1) 7x = or 3x = −

2) 0x = or 8x = −

3) 18x = or 4x = −

4) From line 1 to line 2, the

person forgot to add 25 to

both sides of the equation.

5) 4x = or 10x = −

6) 22x = or 4x = −

7) 1x = or 29x = −

8) 14x = or 4x = −

9) 6x =

10) � = 27 or � = 3

11) � = 22 The two numbers that

satisfy the conditions are 22

and 24.

12) � = 5 The length of each side

of the original garden is 5

yards and the area of the

garden is 25 square yards.

13) � = 2 The ball was in the air

for 2 seconds when it reached

a height of 64 feet in the air.

SECTION 5.2J

1) 7 15x = − ±

2) 9 2 3x = ±

3) 5 4 2x = ±

4) 5 53

2x

− ±=

5) 7 37

2x

±=

6) 14 6 3x = − ±

7) 7 73

4x

±=

8) 1 22

2x

− ±=

9) 1 2

3x

±=

10) When y > 0, there will be 2

real solutions.

When y = 0 there will be 1

real solution.

11) 1.5x = ; The width of the

pool is 15 meters and the

length is 19 meters.

12) 6t = ; The arrow will strike

the ground after 6 seconds.

13) a) ( )2 424h = ; The bottle

rocket is 424 feet high after

2 seconds.

b) 13.13t ≈ or 1.87t ≈ ; The

rocket will be at 400 feet in

the air at two different

times. Once on the way

‘up’ and once on the way

‘down’ during its flight.

SECTION 5.2K

1) 5 12x i= ±

2) 7 2 6x i= ±

3) 13 8 2x i= − ±

4) 6 2x i= − ±

5) 12 6 2x i= ±

6) 1 15

2

ix

− ±=

7) 2 4x i= ±

8) 3 21

10

ix

±=

9) 3 31

8

ix

− ±=

10) 250.81x ≈ ; Emma hits the

golf ball approximately

250.81 yards.

11) 1

2x = or

11

2x = ; At exactly

½ second and 1½ seconds, the

tape measure is at exactly 17

feet above the ground. Gail

can catch the tape measure

anywhere between ½ second

and 1½ seconds after

Veronica tosses the tool.

12) When y < 0 there will be 2

imaginary solutions for a

quadratic equation.

13) Creates the quadratic formula: 2

2

2

2

2 22

2

2 2

2 2

2 2

2

2

2 2

2 2

0

0

0

2 4

4

2 4 4

4

2 4

4

2 2

4 4 or

2 2 2 2

4 4 or

2 2 2

ax bx c

ax bx c

a a

bx cx

a a

bx cx

a a

bx b c bx

a a a a

b b acx

a a a

b b acx

a a

b b acx

a a

b b ac b b acx x

a a a a

b b ac b b ax x

a a a

+ + =

+ +=

+ + =

+ = −

+ + = − +

+ = −

− + =

−+ =

− −+ = − + = +

− −= − − = − +

2

2

2

4

2 2

4

2

c

a

b b acx

a a

b b acx

a

−= − ±

− ± −=



SECTION 5.2L

1) If the quadratic equation does not factor, the roots are either irrational or imaginary. (Answers will vary)

2) a: 2 b: –5 c: –3

x = 3 or 1

2x = −

solve by factoring? Yes, 2 4b ac− (49) is a perfect

square number.

3) a: 1 b: –7 c: 9

7 13

2x

±=

solve by factoring? No, 2 4b ac− (13) is not a perfect

square number.

4) a: 5 b: 3 c: –1

3 29

10x

− ±=


square number.

5) a: 1 b: 1 c: –1

� = −1 ± √52


square number

6) a: 9 b: 6 c: –1

1 2

3x

− ±=

7) a: 2 b: 3 c: –1

3 17

4x

− ±=

8) 1.68 seconds

9) a) 22 seconds

b) 11 seconds

SECTION 5.2M

1) 2 3i

2) 2 6

3) 4i

4) 1x = or 3x = −

5) 3

2x = −

6) 3 2

2

ix

±=

7) 6 11

4

ix

− ±=

8) 1 5x i= − ±

9) 3x = or 5x = −

10) Error on line 6, need to divide

both terms of the numerator

by the denominator value of 2;

3x i= − ±

11) 3 7x = − ±

12) 2x =

13) 13 11

10

ix

±=

14) a) 2.6 seconds

14) b) Since t = 0 means time

when object is initially

thrown, positive time

means time after the

throw was started.

Negative time implies

time before the object was

thrown, which doesn’t

make sense here.

(Answers will vary)

c) The object was at a height

of 25 feet after 0.75

seconds and again at 1.75

seconds.

SECTION 5.2N

1) » Use a graphing utility to find

real zeros.

» Factor and use the zero

product property.

» Square root property

» Complete the square

» Quadratic formula

2) Graph: The solutions are the x-intercepts of x = 3, x = –2

Factor:

( ) ( )3 2 0

3 or 2

x x

x x

− + =

= = −

Quadratic Formula:

1 25

2

1 5

2

3 or 2

x

x

x x

±=

±=

= = −

solutions:

3 or 2x x= = −



SECTION 5.2N (continued)

2) method:: The factoring method

is the fastest and yields rational

solutions with fewer possibilities

of careless errors.

(explanations may vary)

3) Graph: The solutions are the x-

intercepts of x ≈ ±1.73

Square Root:

� − 3 = 0 �� = √3 |�| = ±√3

Quadratic Formula:

0 12

2

2 3

2

3

x

x

x

±=

±=

= ±

solutions: 3x = ±

method: The square root

method. There is no linear term

(the x term). The b-value is equal

to zero. (Answers will vary)

4) a) Graphing calculator (but it

can’t be used to find

complex/imaginary solutions)

b) Completing the Square and

Quadratic Formula

c) Factoring or Square Root

property

5) Factoring – it’s a trinomial that

factors nicely. (method and

explanation may vary)

1

2x = or 3x = −

6) Factoring – it is a binomial with a

common factor. (method and

explanation may vary)

0x = or 1000x =

7) Factoring – it is a trinomial with a

leading coefficient of 1. (method

and explanation may vary)

9x = − or 2x =

8) There are two scenarios for this

situation:

(1) Numbers 8 and 9

(2) Numbers –8 and –9

9) width: 7 in. length: 10 in.

10) Suzie needs 80 feet of fencing.

11) Mike needs 40 feet of fencing.

12) Solutions: 3x = ±

Methods included here may vary

Method 1: Factoring

( ) ( )2 3 3 0x x− + =

Method 2: Square Root Property 2

2

2

2 3 21

2 18

9

continued

x

x

x

+ =

=

=

…

Why? Using square roots has a less chance for careless sign error. Also, we have been using square roots to solve longer than using factoring to solve. (methods may vary)

13) Solutions: 25x = ; so the

width is 25 m and length is 35 m

Method 1:Complete the Square

( )

( )

2

2

2

10 875

10 ____ 875 ____

10 25 900

5 900

continued

x x

x x

x x

x

+ =

+ + = +

+ + =

+ =

…

Note that –35 is extraneous.

Method 2: Factoring

( )

( ) ( )

2

10 875

10 875 0

25 35 0

continued

x x

x x

x x

+ =

+ − =

− + =

…

13) Why? Complete the Square was

more efficient since � !

was a

whole number, and the square

root property is well engrained.

Factoring might have taken

longer to find 2 integers with a

product of –875.

(answers/methods may vary)

14) 3 21

3

ix

− ±= ; solutions should

be verified

15) 6 39

3x

±= ; solutions should

be verified

16) 3x = − (−3) + 6(−3) + 9 = 0

0 = 0

17) 9

ix = ±

81 #�9$

+ 1 = 0

0 = 0 and

81 #−�9 $

+ 1 = 0

0 = 0

18) a) The ball was in the air 5 sec.

b) The ball reaches the max

height at 2.5 seconds.

c) The max height is 100 feet.

d) Graph a second line at y2 = 60

and use “calculate intersect”

to get the x-coordinate that

creates a function value of 60.

The ball is at 60 feet again at

approximately 4.08 seconds.



SECTION 5.2O

(1) – (5) (answers will vary)

Square Root Property:

Eq’n #4: ( )2

3 8x − =

Solution(s): 3 2 2x = ±

��3 + 2√2� − 3! = 8 8 = 8

and

��3 − 2√2� − 3! = 8 8 = 8

Factoring:

Eq’n #2: 2 2 15 0x x− − =

Solution(s): 5 or 3x x= = −

(5) − 2(5) − 15 = 0 0 = 0

and

(−3) − 2(−3) − 15 = 0 0 = 0

Factoring:

Eq’n #3: 2 12 20x x+ =

Solution(s): 6 2 14x = − ±

�−6 + 2√14� + 12�−6 + 2√14� = 20 20 = 20

and

�−6 − 2√14� + 12�−6 − 2√14� = 20 20 = 20

Quadratic Formula:

Eq’n #1: 22 7 15 0x x+ − =

Solution(s): 3

or 52

x x= = −

2 #32$

+ 7 #3

2$ − 15 = 0 0 = 0

And

2(−5) + 7(−5) − 15 = 0 0 = 0

Graphing: Eq’n #5:

( ) 216 24 4755h t t t= − + +

Solution(s): 18.005t ≈

It took approximately 18 seconds for the rock to hit the ground.

−16(18.005) + 24(18.005) + 4755 = 0 0 = 0

6) a) 7 or 1x x= =

b) graph:

c) The intersection points of

the 2 graphs in part b) are

the solutions for x in part a).

9 = (7 − 4) 9 = 9 and

9 = (1 − 4) 9 = 9

7) a) The 4.5 feet is how far the

hose’s nozzle is above the

ground where the water

begins to shoot out.

b) 38.55 feet

c) No. At 27.89 feet from the

nozzle, the stream would hit

the top of the 6 foot fence.

However, at the 28 foot

distance, the water height is

lower, reaching only 5.96

feet high.

8) (answers will vary)

Xmin: –30 Xmax: 30

Ymin: –400 Ymax: 1000

SECTION 5.3A

1) In the quadratic formula,

2 4

2

b b acx

a

− ± −= , the

discriminant is the value of the

expression 2 4b ac− that is

under the radical. This number

is used to determine the

number and type of solutions

of a quadratic equation.

2) discriminant: –4

number of solutions: 2

type: imaginary

3) discriminant: 16


type: real, rational

4) discriminant: 0



5) discriminant: 217


type: real, irrational

6) discriminant: 0





type: imaginary




8) discriminant: 1





type: imaginary

10) the second line of ‘work’, the

negative 6 must be in

parentheses ( ) ( ) ( )2

6 4 1 5− −

discriminant: 16

# of solutions 2


11) a) 240000 0.003 12 27760x x= + +

b) 20.003 12 12240 0x x+ − =

c) 290.88

d) Yes, the discriminant > 0,

which yields two real solutions.

Only the positive solution

would pertain to this story.

12) d = 977. Yes, the discriminant

is greater than zero and yields

two real solutions.

13) d = –280,000. No, since

discriminant < 0 there are no

real solutions.

SECTION 5.3B

1) a) 36

b) 2

c) real, rational

d) � = −2 or � = 4

e)

f) (1, –9)

g) minimum

h) (0, –8)

i) all real numbers

j) ( ≥ −9

2) a) 36

b) 2

c) real, rational

d) � = ±3

e)

f) (0, 9)

g) maximum

h) (0, 9)

i) all reals

j) ( ≤ 9

3) D

4) discriminant = 28

2 solutions, real, irrational

5) discriminant = 0

1 solution, real, rational

6) discriminant = –11

2 imaginary solutions

7) a) negative discriminant

b) zero discriminant

c) positive discriminant

8) Discriminant is a positive

perfect square number

2 real, rational solutions

Possible equation:

( ) ( )2

3 7

10 21

y x x

y x x

= − − −

= − + −

(Answers may vary)

SECTION 5.4A

1) a) Not a solution

b) Solution

2) a) Solution

b) Not a solution

3) a) Not a solution

b) Not a solution

4) (test points may vary)

Use (0, 0)

Is 0 ≤ −4? No, so points

outside the parabola are

solutions.


Use (0, 0)

Is 0 ≥ −3? Yes, so points

outside the parabola are

solutions.





Use (0, 0)

Is 0 > 6? No, so graph

outside the parabola.

7) <

8) ≤

9) >

10) C

11) A

12) F

13) E

14) B

15) D

16)

17)

18)

19)

20)

21)

22)

23)

-4 -2 2 4

-4

-2

2

4

x

y

-2 2 4 6

-4

-2

2

4

x

y

-6 -4 -2 2

-2

2

4

6

8

x

y

-6 -4 -2 2

-2

2

4

6

8

x

y

-10 -8 -6 -4 -2 2

-6

-4

-2

2

4

x

y

-2 2 4 6

-4

-2

2

4

x

y

-6 -4 -2 2

-2

2

4

6

8

x

y



SECTION 5.4B

1) a) When � < −4or� > 2 b) −4 < � < 2 c) graph

2) a) −3 < � < 2 b) � < −3or� > 2 c) graph

3) Answers will vary, but

x-intercepts and concavity must match. Example:

4) Answers will vary, but

x-intercepts and concavity must match. Example:

5) � < −5or� > 5

6) � ≤ 0or� ≥ 4

7) 2 < � < 5

8) 5 73 5 73

2 2x

− +≤ ≤

9) 2 8 20 0x x− − > 2 or 10x x< − >

10) 2 12 27 0x x− + < 3 < � < 9

11) � ≤0

or� ≥ 5

12) 2

53

x< <

-5 8

-5

5

x

y

Function values are

LESS than 0 below x-axis

Function values are


Function values are

MORE than 0 above x-axis

-5 5 10

-5

5

x

y

Function values are


Function values are


Function values are


Function values are


Function values are




SECTION 5.4 B (continued)

13) $20 < � < $100 (line is

dashed)

x-intercepts are 20, 100 and

vertex is at (60, 1600)

14) 1 ≤ � ≤ 3 seconds

x-intercepts are at 1 and 3,

vertex is at (2, 16)

15) a) 0 ≤ � ≤ 8.03ftand � ≥ 18.45ft. The

x-intercepts are at 8.03

and 18.45 and the vertex

is at (13.24, 1.47)

b) No. From 15 feet away,

the height of the ball

would be 9.3 feet, so the

ball would go over the top

of the net. Also, from the

graph and algebraic

solution inequalities

above, 15 is not in the

correct ranges

(explanations may vary).

16) 4 ≥ 2 inches. Values in the

range 4 ≤ −2 are extraneous.

x-intercepts are at 2 and –2,

the vertex is at (0, -5920)

17) Driver’s age is between 56.6

years and 70 years (inclusive

on 70).

Unit 5 – Review Material

1) Answers will vary a) There is no x-term, such as

� = 4orif it is in vertex

form, like ( )2

3 2 5 0x − + =

b) If it is unfactorable and a = 1 and b is even c) a, b, c are all integers and

are fairly small numbers

d) If it is unfactorable and a, b, and c are larger numbers and/or decimals.

2) Answers may vary Suggest A, J and K for square

root method Suggest B, F and I for factoring Suggest C, D, and E for completing the square Suggest G, H and L for quadratic formula

3) a) � = ±3√2

b) � = 1 ± 2�√3

c) � = 13or� = −1

4) a) � = −10 ± 2� b) � = 3 ± 2√6

c) � = 2or� = −6

5)a)� = −2

3or� = −5

b)� = −3

2or� = 5

c)� = −2

3

6)a)� =−3 ± √3

3

b)� =4 ± �√2

6

c)� = 2or� =3

5

7)a)� = −4 ± 5√2

b)� = 1or� = −1

2

c)� = −2 ± �√5

d)� =7 ± 2�√3

14

8) t = 5 seconds

9) 5.6 seconds

10) a) Disc. is negative. There

are two imaginary

solutions.

b) Disc. is zero. There is one

real, rational solution.

c) Disc. is positive. There are

two real solutions.



Unit 5 – Review Material (continued)

11) a) Disc. is –16. There are

two imaginary solutions.

b) Disc. is 261. There are

two irrational solutions.

c) Disc. is 0. There is one

real, rational solution.

12) B

13) −5 < � < −1

14)

• Algebraically find the

x-intercepts.

• Sketch the graph of a

parabola that has these

x-intercepts and opens up if

a > 0 or down if a < 0. Also

determine dashed or solid

line.

• Identify the x-values for

which the graph lies below

the x-axis (#15a) or above

(or on) the x-axis (#15b).

• For ≤ or ≥ include the

x-intercepts in the solution.

15) a) −2 < � < 1 Function

values less than 0 (below

x-axis)

b) � ≤ −8or� ≥ −4

Function values are ≥ 0 (on

or above the x-axis)

16) Graph B

17) � ≤ −1or� ≥ 4



SECTION 6.1A

1. a) x < –1.69, x > 2.36

b) –1.69 < x < 2.36

c) (–1.69, 12.6) Relative Max

( 2.36, –20.75) Relative Min

These occur at turning points

d) (–3, 0), (0, 0), (4, 0)

e) (0, 0)

f) none

2. a) i) 1 ii) –1 iii) –1 iv) 1

b) If a is positive, the ends

have a positive slope ��

If a is negative, the ends

have a negative slope �

c) They all either look like

or . Some have

more of a squiggle in the

middle, with relative

maximums and minimums.

They differ in the number of

times each graph crosses the

x-axis (answers may vary).

3. a) They are all using different

windows, so they are seeing

different portions of the

graph. (answers may vary)

b) Meng’s is the best because

you can see all the key

features: max, min,

intercepts, end behavior.

(answers may vary)

4. Sign: Positive Domain:ℝ

Range: ℝ Rel Min: approx. (0.5, –8.5) Rel Max: (–5.5, 2) Increasing: x < –5.5 and x > 0.5 Decreasing: –5.5 < x < 0.5 x-int(s): (–7, 0), (–4, 0), (3, 0) y-int: approx. (0, –8.33) 5. Sign: Negative Domain:ℝ

Range: ℝ Rel Min: approx. (–4.1, –1.1) Rel Max: (0, 6) Increasing: –4.1 < x < 0 Decreasing: x < –4.1 and x > 0 x-int(s): (2, 0), (–3, 0), (–5, 0) y-int: approx. (0, 6)

6. Sign: Positive

Domain:ℝ

Range: ℝ

Rel Min: approx. none

Rel Max: none

Increasing: ℝ

Decreasing: none

x-int(s): (0, 0)

y-int: approx. (0, 0)

7. Table

# #1 #3 #5

x= (–5,0)

(3,0)

(7,0)

(0,0)

(3,0) (1,0)

y= (0,-5.5) (0,0) (0,1)

Max (5.2, 2) (1,4) None

Min (–1.8, –7) (3,0) None

Inc –1.8<x<5.2

x<1

x>3 None

Dec x < –1.8

x > 5.2 1<x<3

Dec. over

entire dom.

Continued…

#4 #6 #2

(0,0) (–2.4, 0)

(–8, 0)

(–2, 0)

(9, 0)

(0,0) (0, 4) (0, 3.7)

None (–1.2, 7) (4.5, 9.5)

None (1.2, 0.9) (–5.5, –3.2)

Inc. over

entire dom.

x > 1.2

x < –1.2 –5.5<x<4.5

None –1.2<x<1.2 x < –5.5

x > 4.5

8. Explanations will vary.

a) Yes – if before year 1993

the number of acres was less

than 74,630 and declining

each previous year, then the

end behavior would be (↙↗)

b. Yes – if sometime after

1996 the number of bales

was > 9400 thousand and

increasing each succeeding

year, the end behavior

would then be (↙↗).




8. c) Yes – if before the year

1993 the yield per acre

was < 5.22 thousand and

decreasing each previous year,

the end behavior would then

be (↙↗).

9. Answers will vary but will all

be of the form y = x3.

10. Answers will vary but will all

be of the form y = –x3.

SECTION 6.1B

1. Sign: Positive

End: (� �)

Domain: ℝ

Range:ℝ

Rel Min: (1.29, –1.30)

Rel Max: (–1.29, 7.30)

Inc: x < –1.29 and x > 1.29

Dec: –1.29 < x < 1.29

x-int(s):(–2.49, 0),(0.66, 0),(1.83, 0)

y-int(s): (0, 3)

2. Sign: –Negative

End: ( �)

Domain: ℝ

Range:ℝ

Rel Min: none

Rel Max: none

Inc: none

Dec: ℝ

x-int(s): (0.538, 0)

y-int(s): (0, 3)

3. Explanations may vary.

a determines the end behavior. If

a > 0 the end behavior is (� �) and if a < 0 the end behavior is

( �). d is the y-intercept.

4. Sign: Positive

End: (� �)

Domain: ℝ

Range:ℝ

Rel Min: (2.08, –1.21)

Rel Max: (–2.08, 6.01)

Inc: x < –2.08 and x > 2.08

Dec: –2.08 < x < 2.08

x-int(s): (–4, 0), (1, 0), (3, 0)

y-int(s): (0, 2.4)

5. Sign: –negative

End: ( �)

Domain: ℝ

Range:ℝ

Rel Min: (–4.46, –8.31)

Rel Max: (2.46, 8.31)

Inc: –4.46 < x < 2.46

Dec: x < –4.46 and x > 2.46

x-int(s): (5, 0), (–7, 0), (–1, 0)

y-int(s): (0, 3.5)


The x-intercepts occur at m, n,

and p. The y-intercept is

:(−;)(−�)(−<). a

determines the end behavior.

If a > 0 the end behavior is

(� �), and if a < 0 the end

behavior is ( �).

7. Sign: Positive

End: (� �)

Domain: ℝ

Range:ℝ

Rel Min: none

Rel Max: none

Inc: ℝ

Dec: none

x-int(s): (–1.41, 0)

y-int(s): (0, 23)

8. Sign: Negative

End: ( �)

Domain: ℝ

Range:ℝ

Rel Min: none

Rel Max: none

Inc: none

Dec: ℝ

x-int(s): (3.41, 0)

y-int(s): (0, 60.5)


k moves the graph up and down;

h moves the graph left and right.

a stretches the graph. If a is

negative, the graph always

slopes down. If a is positive

the graph always slopes up.

There will be no maximum or

minimum values.

10) Sign: Positive

End: ( �)

Domain: ℝ

Range: ( ≥ −4.3

Rel. Min. (–1.9, –4.3), (1.1, 1.95)

Rel. Max. (0, 4)

Inc: −1.9 < � < 0, � > 1.1

Dec: � < −1.9, 0 < � < 1.1

x-int: (–2.4, 0), (–1, 0)

y-int: (0, 4)



SECTION 6.1B (continued)

11) Sign: Positive

End: (↙↗) Domain: ℝ

Range: ℝ

Rel. Min. (–0.23, –2.03), (4.5, –2.6)

Rel. Max. (–4.1, 7.3), (2.2, –0.7)

Inc: � < −4.1, −0.23 < � < 2.2, � > 4.5

Dec: −4.1 < � < −0.23, 2.2 < � < 4.5

x-int: (–5.3, 0), (–1.7, 0), (5.5, 0)

y-int: (0, –2)

12) Sign: Positive

End: (↖↗) Domain: ℝ

Range: ( ≥ −5

Rel. Min: (3, –5)

Rel. Max: None

Inc: x > 3

Dec: x < 3

x-int: (0.8, 0), (5.2, 0)

y-int: (0, 11.2)

13) Sign: Positive

End: (↖↗) Domain: ℝ

Range: ( ≥ −8.96

Rel. Min: (–2.45, –6.21),

(1.29, –8.96)

Rel. Max: (–1, 0)

Inc: –2.45 < x < –1, x > 1.29

Dec: x < –2.45, –1 < x < 1.29

x-int: (–3, 0), (–1, 0), (4, 0)

y-int: (0, –3.84)

14) Explanations may vary.

Even degree: end behavior is

either (↖↗) or (↙↘). Odd degree: end behavior is

either (↙↗) or (↖↘). 15) a) 4272.9 million ft2 when

t = 18, since 1990 is 18

years after 1972.

b) Increasing. Explanations

may vary, but graph is

always increasing on this

interval.

c) No. When 0 < t < 24, S

increases each year.

15) d) Domain: 0 ≤ � ≤ 24

where t is the number of

years since 1972.

Range: 1700 ≤ @ ≤ 5050.5

S is the retail space (in

millions of ft2) over that

time period.

16) a) Domain: 0 ≤ � ≤ 23,

Number of years from 1980

to 2003.

Range: $7.70 ≤ A ≤ $30.99,

monthly rate for cable TV in

that time period.

b) In the year 2003, the

maximum rate was $31.06

(23.8, 31.06). After that, the

rates decrease over time…

Not the case! For 2014 this

model would yield a rate of

$14 per month!

c) 0 ≤ � ≤ 23, cable TV rates

were always increasing

from 1980 to 2003.

d) When t > 23.8, indicating

the rates started declining in

the year 2003.

e) A(3) = $8.55

17) Explanations may vary

a) L(18) = 50.7 in.

H(18) = 52.3 in.

50.7” < normal height < 52.3”

b) No. After a certain age, a

heifer does not get any

taller.

c) 6.6 mo. < age < 8.3 mo.

Draw a horizontal line

y = 43 and use “Calc.

Intersect” on both curves

L(8.3) = 43 and H(6.6) = 43.

d) At the point of inflection,

the increasing height starts

to level off before the graph

starts to increase again.

This occurs when the height

is near 55", so probable age

would be

30 mo. < age < 31.6 mo.


a) Domain: 0 < x < 6 Width > 0 and 12 – 2x > 0 ⟹ � < 6

Range: 0 < V < 194.07

Volume > 0, and within

given domain the max.

volume is 194.07

b) 194 in3 within the

acceptable domain

0 < x < 6. The max volume

occurs at (2.26, 194.07)

c) Approximately 2.25 in2


a) y-int. = 10. The roller

coaster is 10 ft. above ground

before it starts a certain

portion of the track.

b) Yes. (20, 42) => after 20

seconds the roller coaster

reaches its maximum height

of 42 feet.

c) Yes. (60, 10) => After 60

seconds the roller coaster

returns to a height of 10 feet.

d) No. The height never

equals 0 after the coaster

begins rolling.

e) H(5) = 25.1. After 5

seconds the roller coaster has

reached a height of 25.1 feet

above the ground.

f) At t = 10 seconds and at t =

32.1 seconds. The roller

coaster’s height of 35 feet is

obtained twice during the ride

– once on the way up and

once again on its way down.

At t = 77 the height is again

35 feet, but the roller coaster

ride lasts 60 seconds. 77

seconds is not part of the

domain.

g) After 60 seconds the

height of the coaster is

continually increasing to ∞.



SECTION 6.1B

20) a) 4 mg

b) 2 days

c) Domain: 0 ≤ � ≤ 2 after 2 days the drug is completely out of the patient’s bloodstream (y = 0)

Range: 0 ≤ D(�) ≤ 8. 8mg is the original amount of the drug in the patient’s bloodstream which lessens

each hour afterwards. (Explanations for 20c may vary)

SECTION 6.2A

1) C

2) A

3) B

4) B

5) A

6) B

7) a) True

b) False: 6� + 9� = 15�

c) False: (2�EFG)H = 8�0EF0

8) True: Using the power of a

power rule, �I = � I. By the

commutative property of

multiplication, ab = ba.

9) False: using power of a power

rule (�)H = �∙H = �K

10) 4�E

11) 10L;

12) 24�0K

13) 9�M(0N

14) 16;0K

15) 8�O

SECTION 6.2B

1) Jamal added −3�and − 5�, which are not like terms. The

correct answer is

9�H − 3� − 5� − 2

Precious also added unlike

terms 5�and − 4�. She had a

second error when she added

−3�H + 8�Htoget5�K. Those

exponents don’t combine. The

correct answer is

5�H + 5� + 4� − 5.

Kiarra used the distributive

property incorrectly.

– (– 8� + 3� − 7) = 8� − 3� + 7.

The correct answer is

12� − 12� + 7

2) 7x + 1

3) 2< + 5< − 4

4) 10� − 7

5) 2�

6) 7: − 5:� − 3:H

7) 2� − 9� + 3

8) 10� − 5( + 10

9) 6;� + 8;� − 28

10) 7� + 2� − 3

11) a + 7b

12) 5x + 12z

13) p – 8q

14) 7:� − 2:� + 2�

15) −6�H + 6�( + 3�(

16) D = 6� − 12� + 9

17) � − 2� + 13

18) x + y

19) 2x + 6y

20) �G + �H − 10� + 3

21) 5� − 10� meters

22) S(D) = 0.5< + 78< + 12,650

23) −11� + 10

24) 14� + 6� + 8

25) a) 2

b) 5x

c) 8

d) 7x

e) –1

SECTION 6.2C


Thao used the product property

of powers incorrectly. He

multiplied � and −3�H to get

−3�K. It should be −3�E.

Similarly, he multiplied � and

5� to get 5�. It should be 5�H.


−3�E + 26�H − 4� − 35� + 28

Louis mixed up the +/– symbols

when combining 7x and –3x to

get –10x instead of 4x. The

correct answer is � + 4� − 21.

Monique didn’t distribute

the –3 over the last two terms.


−6� + 15� − 21. 2) 4�H − 20� + 32�

3) −7�E + 13�H − 20�

4) 3�G − 27�H + 47� − 21� − 98

5) �H + 7� − 5� − 35

6) 6�H − 8� − 12� + 16

7) 3�H + 2� − 51� + 70

8) 36� + 60� + 25

9) 5�H − 34� + 99� − 60

10) 6�K − 2�E + 24�G − 20�H

11) 8�E − 6�G + 25�H − 12� + 3� − 18

12) ( = 3�E + 9�G + 2�H + 6� − 8� − 24

y-int. = –24

13) ( = 9� − 42� + 49

y-int. = 49

14) ( = −24�H + 180� + 6� − 45

y-int. = –45

15) ( = �G − 4�H + 12� − 4� + 11

y-int. = 11

16) ( = −�H − 3� + 8� + 24

y-int. = 24

17) ( = 5�H − 38� + 76� − 33

y-int. = –33



SECTION 6.2C (continued)


a) The end behavior is

determined by looking at

the degree (the highest

exponent, which should

be the first term) and the

leading coefficient.

b) The y-intercept is the

constant, or last term in

standard form.


a) 78.45; In 1995 the average

amount of bananas (in

pounds) eaten per person in

the US was 78.5.

b) Increasing pounds of

bananas consumed per

person from 1995 to 1996,

then decreasing from 1996

to 2000.

20) a)

( ) 3 20.427 5.445 346.517 3679.92R t t t t= + + +

b) $3,679,920,000

c) y-intercept

21) Width = 15 feet

Length = 30 feet

22) a)

( ) 3 240 6000 240, 000R t x x x= − + + −

b) A(0) = $ − 240,000

c) Yes; negative number for

revenue means the costs

were higher than the sales

(which was $0)

(explanations may vary)

23) T = 6� − 2� − 20

24) 20�H − 28� + 15� − 21

25) 10� − 3� − 27

26) T = 53� + 16� − 3

27)

( )

( )

3 7 30 3

3

3 21 30

3

3 7 10

3

7 10 3

3 3

nn

nn

nn

n n

− +− =

− +− =

− +− =

− + − =

=

28) a) 3

b) 2x

c) –7

d) 3

29) �H + 3� + 3� + 1

30) �H + 3� + 2�


The number of bags of various

flavored chips times the price

per bag gives the amount of

money collected from the

sales.

SECTION 6.2D

1) 3 22 3 3 7x x x− − +

Yes, (x + 9) is a factor.

2) 4 3 2 35 7 16 2 3

2x x x x

x+ + + − −

−

No, (x – 2) is not a factor.

3) 31

2 62 6

xx

− ++

No, (2x + 6) is not a factor.

4) 5 4 3 2 24 4 4 9 6 5

1x x x x x

x− − − − − − +

−


5) 2 2 2x x− +

Yes, (2x – 1) is a factor.

6) 2 124 1

2x x

x+ − +

+

No, (x + 2) is not a factor.

7) 24 17 16x x+ +

Yes, (x – 8) is a factor.

8) 4 3 2 43 3 7 7 6

1x x x x

x+ + + + +

−


9) 2 2 1x x+ −

Yes, (x – 5) is a factor.

10) #7 and #9; the divisor is a

factor of the dividend.

11) Lisa is correct, except to write

the quotient:

2 54 2

2x x

x+ + −

−

Maut is missing a place

holder for the linear (x) term.

The correct quotient for his

problem: 3 210 5 9 9x x x− + −

Craig is used “–2” for the

outside value and should have

used a value of “2” (the value

that creates a zero value for

the divisor. The correct

quotient for his problem:

12

2x

x− −

−

12) ( )2 5x − is the length of the

rectangular garden.

13) ( )22 4 6x x+ − is the base of

the triangle.

14) ( )4x − is the width of the

mural.

15) a) ( )2x +

b) B is the divisor, located in

the denominator of the

remainder fraction.

c) 3 25 3 21 8A x x x= − + −

16) One example would be

( ) ( )32 3 1x x x− + ÷ −

17) One example would be:

( ) ( )3 2 23 3 1x x x x− + − ÷ −

Synthetic division can only

be used when the divisor is a

binomial with degree 1, like

( )x c−



SECTION 6.2E

1) a) There are 3 distinct linear factors, which yield 3 x-intercepts at

( ) ( ) ( )12, 0 5, 0 , and 8, 0− − .

b) There is one duplicate linear factor which gives a double root at x = –5 (but only one x-intercept here) and the other linear factor yields the x-intercept at

( )1, 0 .

c) There are 3 distinct linear factors which yield 3 x-intercepts at

( ) ( )1

, 0 6, 0 , and 10, 02

−

2) zero, root, solution 3) algebraically: set the

function equal to zero (0) and solve for x.

graphically: look for the x-intercept(s) and/or use a graphing calculator to “calculate the zeros”.

4) No; there is a remainder value different than zero.

5) No; there is a remainder of 2. If (x – 1) was a factor, there would be a zero value for a remainder.

6) Yes; because the remainder is zero, this means that (x + 4) is a factor of the polynomial, which means x = –4 is a zero (or solution).

7) No; the remainder is 5 (not zero) so x = 3 is not a zero.

8) k = –4 9) k = 0

10) ( ) ( ) ( )2 2 5f x x x x= − +

11) ( ) 3 26 4 24P x x x x= − − +

12) a) ( ) ( ) ( )5 2 1y x x x= − + +

b) ( ) ( ) ( )1 5 5y x x x= + − −

c) ( ) ( )2

2 1y x x= + −

d) ( ) ( ) ( )1 5 5y x x x= + + −

13) a) ( ) ( ) ( )2 1 3y x x x= − + +

b) ( ) ( ) ( )4 3 5y x x x= + − −

c) ( ) ( ) ( )9 1 1 2y x x x= + − +

d) ( ) ( ) ( )2 1 5 4y x x x= + + −

e) ( ) ( ) ( )3 2 5 2y x x x= − + +

f) ( ) ( ) ( )5 1 1 3y x x x= + − +

14) a) Height: ( )3 10x −

Width: ( )1x +

b) length: ( )2 5x +

height: ( )5x +

SECTION 6.3A

1) a) Real zeros: x = –10, –1, 3

Factors: ( ) ( ) ( )10 1 3x x x+ + −

Possible equation:

( ) ( ) ( )1

10 1 320

y x x x= + + −

b) Real zeros:

x = –3(double root), 2

Factors: ( ) ( ) ( )3 3 2x x x+ + −

Possible equation:

( ) ( ) ( )1

3 3 25

y x x x= − + + −

c) Real zeros:

x = –7, –2, 1

Factors: ( ) ( ) ( )7 2 1x x x+ + −

Possible equation:

( ) ( ) ( )1

7 2 15

y x x x= + + −

2) a) ( ) ( ) ( )1, 0 , 3, 0 , 4, 0

b) ( ) ( ) ( )4.879, 0 , 1.287, 0 , 1.592, 0−

c) ( ) ( ) ( )0.438, 0 , 4.562, 0 , 9, 0

d) ( )3, 0−

3) a) 1

3x =

b) 2.059, 0.469, or 0.777x = − −

c) 2,1, or 3x = −

d) 1, or 2x = −

4) Height: 8 feet

Width: 15 feet

Length: 40 feet

5) Height: 8.85 feet

Width: 6.85 feet

Length: 13.85 feet

6) They will need to sell 30 cars

7) x = 0 meters or x = 6 meters

SECTION 6.3B

1) a) Yes; Jebediah knows he

can find the x-intercepts

(solutions) where the

y-value equals zero.

Kalani knows she can look

at the intersection(s) of her

2 graphs to find the

x-values (solutions) where

y = 8.

b) Yes; both are effectively

solving a system of two

equivalent equations by

graphing.

3 2

1

2

0.5 2.5 0.5 2.5Jebediah

0

y x x x

y

= − + + −

=

3 2

1

2

0.5 2.5 0.5 5.5Kalani

8

y x x x

y

= − + + +

=

2) x = –3, –1, 1

3) x = –3, –1.5, 3



4) x = 2

5) a) 2.607,1.392, or 7.715x = −

b) 2.574, 0.177, or 4.398x = −

c) 4.927x =

d) 7.786, 1.410 or 6.196x = − −

6) a) 2.067 or 1.284x = −

b) 1.446, 0.177, or 2.086x = −

c) 4.511, 0, 0.759 or 1.753x = −

d) 6.169, 3.623, 1.230 or 7.022x = − − −

7) There are 3 situations that

demonstrate there are a

total of 3 solutions:

� 1real solution, 2 complex

solutions (show a graph)

� 2 real solutions (1 of these

being a double root), 0

complex solutions (show a

graph)

� 3 real solutions , 0

complex(show a graph)

8) a) Graph:

b) In the year 1996 (1.7 years

after 1995).

c) The average number of

pounds equals 13.5 pounds

three different times:

When x ≈ 1.2 ⇒ year 1996

When x ≈ 2.7 ⇒ year 1997

When x ≈ 5.3 ⇒ year 2000

9) a) 6.3%

b) 26%

c) 47%

10) a) 361 arrests; it is the

y-intercept

b) 14.2 years after, so in the

year 2004

c) In the year 2010

d) About 22 years after 1990,

so in the year 2012; this is

represented by the

x-intercept

11) a) In the year 2034 and again

in the year 2108

b) 189.3 million; this is a

relative maximum for time

after 1970.

c) When x = 104.3 years after

1970, which is in the year

2074

12) 1989

13) a) ( ) ( ) ( )2, 6 , 1, 5 , 1, 3− − −

b) method 1: using the

graphing method as shown

here, the solutions are the

x-coordinates of the

point(s) of intersection, so

x = –2, –1, 1

method 2: solve this

algebraically. Using

substitution, set the two

expressions equal to each

other:

( ) ( ) ( )

1 2

3 2 2

3 2

5 4 2

2 2 0

1 1 2 0

1, 1, 2

y y

x x x x x

x x x

x x x

x x x

=

+ − = − − +

+ − − =

+ − + =

= − = = −

SECTION 6.3C

1) a) 5 3 3

2,2

x− ±

= −

b) 5, 3 3x i= − ±

c) 0, 5, 1 11x = − ±

2) a) 5,1 33x = ±

b) 1

, 3 2 33

x = − − ±

c) 0, 3, 3 3x i= − − ±

3) a) 1 5

2, ,3 2

x = −

b) 1, 3 7x = − − ±

c) 3, 1, 2 2x = − − ±

4) a) 5 17

2,4

x±

= −

b) 3, 1 2x i= − ±

c) 1 1 3

,2 2

ix

− ±= −

5) a) 3 5

2,2

x− ±

=

b) 2, 4,1 7x i= − ±

c) 3 (double root), 2 3x = − ±

6) a) 5, 2, 0.75x = −

b) 4,1.5,10x = −

c) 2, 1 3x i= − ±

d) 3, 0, 7x = −

e) 2,1, 2x = −

f) 3, 3x i= ±

7) A cubic function always has 3 solutions: �Either 1 real and 2 imaginary solutions (show graph with 1 x-intercept) �2 distinct real solutions (with one of them being a double root – creating three solutions in total) (show graph with curve that crosses over the x-axis and one section that has a vertex that touches the x-axis). �3 real solutions (all different) (show graph with curve crossing over the x-axis three distinct times).



Unit 6 – Review Material

1) a) Standard Form

b) It is the y-intercept

c) x y

–3 –20

–2 0

–1 6

0 4

1 0

2 0

3 10

d) Rel min: ( )1.54, 0.88−

Rel max: ( )0.87, 6.06−

Domain: all real numbers

Range: all real numbers

Inc. int.: 0.87, 1.54x x< − >

Decr. int.: 0.87 1.54x− < <

Zero(s): ( ) ( ) ( )2, 0 , 1, 0 , 2, 0−

2) a) 23 12 7y x x= − + −

b) 3 22 43 140y x x x= + − −

3) a) Intercept form or Factored

form

b) It brings forward the

information for finding the

x-intercept

c) x y

–3 12

–2 0

–1 –2

0 0

1 0

2 –8

3 –30

d) Rel min: ( )1.22, 2.11− −

Rel max: ( )0.55, 0.63



Inc. int.: 1.22 0.55x− < <

Decr. int.: 1.22, 0.55x x< − >

Zero(s): ( ) ( ) ( )0, 0 , 1, 0 , 2, 0−

y-intercept: ( )0, 0

4) a) It is similar to the vertex

form of a quadratic function,

but this is a cubic function.

b) It provides information about

the inflection point.

c) x y

–2 –9.5

–1 0

0 3.5

1 4

2 4.5

3 8

d) Rel min: none

Rel max: none



Inc. int.: x−∞ < < ∞ , or ℝ

Decr. int.: none

x-intercept: ( )1, 0−

y-intercept: ( )0, 3.5

5) a) 3 214 2 3 6x x x+ + −

b) 3 25 10 18x x− +

c) 22 5 3x x+ +

d) 3 23 16 17 30x x x+ − −

e) 3 22 24 96 123x x x+ + +

f) 2 6 3x x− +

g) 3 2 705 18 36

2x x x

x− + − +

+

6) a) During month #5

b) 100 coats

c) 50 coats

d) x > 5.5

e) 0 < x < 5.5

f) no

7) a) ( ) ( ) ( ) ( )1 1 7f x x x x= + − −

b) ( ) ( ) ( ) ( )1 4 3f x x x x= − − +

8) a) 5, 3, 3x = − −

b) 4

5, , 23

x = − −

c) 1 1

0, ,2 3

x = −

d) 3, 2,1x = − −

[8(a) – 8(d) the procedure for

finding the rational zeros may

vary.]

9) a) 1

, 4, 32

x = −

b) 1

, 6, 22

x = −

10) 1 3

3, 5,2

ix

− ±= −

11) a) 1 11

2, 1,2

ix

− ±= − −

b) 3 5

3, 4,2

x±

=



SECTION 7.1A

1. a)

• From the store, drive

South on Aspen St

• Turn Right on Elm St

• Turn Left on Acorn St

• Go past one stop sign and

turn Right/West on Hwy 27

• Travel 8mi to your home

b) I drew a picture of the route

TO the store and then followed

it backwards.

c) Reverse

2. a) 1

b) 15

c)U(�) = (VWH)X0

E

d)Y(�) = EVW0

+ 3

e) An inverse function is the

reverse process of a

function; it “undoes” what a

function “does.”

3. a)

Pairs of Skates Additional Pay

4 $20

20 $100

8 $40

13 $65

x $5x Z

[

$x

b) The extra pay P(x) is 5

times the # skate pairs

sharpened x.

c) The number of pairs of

skates N(x) is0

E the extra

pay x.

d) P(3) = 15; If he sharpens 3

pairs of skates, he will

make $15 additional pay.

e) N(15) = 3; If he made $15

additional pay, he must

have sharpened 3 pairs of

skates.

f) They give the same results,

backward.

g) i)

x P(x) x N(x)

1 5 5 1

2 10 10 2

3 15 15 3

4 20 20 4

ii) They are reversed.

4. a) b)

x f(x) x f -1

(x)

–3 4 4 –3

1 –2 –2 1

5 2 2 5

c)

d) No – Each input (x) does

not have exactly one

output (y).

[ex. (0, 3) and (0, –0.5)]

5. a) {(5, –13), (–9, –9), (–4, 0), (6, 4),

(10, 9)}

b) Yes – Each input (x) has

only one output (y).

6. a) {(4, –2),(7, 4),(11, 0),(7, –3)}

b) No – the input 7 has two

different outputs (4 and –3)

7. a)

x –2 –1 0 1 2

g-1

(x) –10 –6 –2 2 6

b)

c) Yes – Each input (x) has

only one output (y).

8.

x y = 2x x y = 0

�

–3 –6 –6 –3

–2 –4 –4 –2

–1 –2 –2 –1

0 0 0 0

1 2 2 1

2 4 4 2

3 6 6 3

Yes – the inputs (x) and

outputs (y) are reversed.

9.

x y = x2 x y = √�

0 0 0 0

1 1 1 1

2 4 4 2

3 9 9 3

4 16 16 4

5 25 25 5

6 36 36 6

7 49 49 7

-5 5

-5

5

x

y

-5 5

-5

5

x

y



Yes – the inputs (x) and

outputs (y) are reversed.

10. a) 770 pounds

b) It would tell us how much

money (in US dollars) we

brought with us, given we

received a known amount

of British pounds.

c) 4 = \0.EG

d) $191.56

11. -Choose points on the graph.

-Reverse the coordinates

(x, y) � (y, x)

-Graph the new coordinates

SECTION 7.1B

1. a) Values lower than 3 would

make a negative in the √

b) The smallest √ result would

be 0, and then adding 5

gives a minimum value of 5,

so y cannot be less than 5.

c) Think about (guess/check)

what #s can’t work for x and

then #s that won’t be results

for y. (see a) and b))

d) There are no points graphed

with an x-value before 3;

there are no points graphed

with a y-value less than 5.

e) There are “error”s before x = 3

and no y-values below y = 5.

2. a) Equation

Domain: x > 7

Range: y < –4

Explanation: an x-value

lower than 7 would make a

neg. under the √; the greatest

√ result is 0, and then

subtracting 4 gives a max

value of –4, so y cannot be

greater than –4.

b) Graph

Domain: x > –8

Range: y > –7

Explanation: There are no

points graphed with an

x-value before –8; there are

no points graphed with a

y-value below 7.

2. c) Table of values

Domain: x > 3

Range: y > 5

Explanation: There are

“error”s before x = 3 and no

y-values less than 5.

3. Milo is correct – Basra appears

to be thinking about the

range.

4. [B]

5. [C]

6. [A]

7.

Increasing

Domain: x > 0

Range: y > 2

x-int: none y-int: (0, 2)

8.

Decreasing

Domain: x > –4

Range: y < 3

x-int: (5, 0) y-int: (0, 1)

9.

Decreasing

Domain: x > 0

-4 -3 -2 -1 1 2 3 4

-5

-4

-3

-2

-1

1

2

3

4

5

x

y

-4 -3 -2 -1 1 2 3 4

-5

-4

-3

-2

-1

1

2

3

4

5

x

y

-4 -3 -2 -1 1 2 3 4

-8

-7

-6

-5

-4

-3

-2

-1

1

2

x

y



Range: y < –2

x-int: none y-int: (0, –2)

10.

Increasing

Domain: x > 0

Range: y > 0

x-int: (0,0) y-int: (0,0)

11.

Increasing

Domain: x > 1

Range: y > 2

x-int: none y-int: none

12.

Increasing

Domain: x > –5

Range: y > –4

x-int: (–1, 0) y-int: (0, 0.47214)

13. Shifted Left 3 and Down 4

14. Flipped

Shifted Right 2 and Up 515.

Graph Number 2 3 1 4

Beginning Point (0,0) (-6,0) (2, -4) (-2,4)

x-intercept (0,0) (-6,0)

y-intercept (0,0) (0,2.6) None (0,2.6)

Inc./Dec. Dec. Inc. Inc. Dec.

Domain x > 0 x > -6 x > 2 x > -2

Range y < 0 y > 0 y > -4 y < 4

16. Domain: x > –3 Range: y > –4

17. Domain: x > 4 Range: y < 1

18. Answers may vary

a) ( � √� � 5 � 2

b) ( � �√� � 7 � 10

c) ( � √� � 9

SECTION 7.1C

1. a) No – you can ∛ any number

(pos or neg).

b) No – since x can be anything

and √�^ could be anything,

adding 6 could lead to any #.

c) (–3, 6)

d) The x-coordinate is the

result of setting what is under

the radical equal to zero and

then solving for x; the

y-coordinate is the value of

the constant added/subtracted

outside the ∛. e) Increasing – it is a positive ∛.

2. a) Increasing

Domain: 7

Range: 7

Point of Inflection: (2, –4)

b) Decreasing

Domain: 7

Range: 7

Point of Inflection: (3, –5)

3. Neither – should be (12, 10)

Arturo didn’t set what is under

the radical equal to zero and

then solving for x.

Kira has the x and y switched.

4.

Decreasing

Domain: 7

Range: 7

Point of Inflection: (0, 0)

-8 -4 4 8

-8

-4

4

8

x

y



5.

Increasing

Domain: ℝ

Range: ℝ

Point of Inflection: (–4, 1)

6.

Increasing

Domain: ℝ

Range: ℝ

Point of Inflection: (0, 4)

7.

Decreasing

Domain: ℝ

Range: ℝ

Point of Inflection: (–3, 5)

8.

Decreasing

Domain: ℝ

Range: ℝ

Point of Inflection: (–3, –2)

9.

Increasing

Domain: ℝ

Range: ℝ

Point of Inflection: (–5, –1)

10. [C]

11. [A]

12. [B]

13. a) Answers will vary:

Graph: Identify where the

curve changes from a hill to a

bowl.

Table: Look for a set of three

y-values that are the same

distance apart; the inflection point is the middle of these three.

Equation: The x-coord. is

calculated by setting what is

under the radical equal to zero

and then solving for x; the

y-coordinate is the number

added/subtracted outside the ∛. b) Answers will vary:

Graph: Look left to right to see

if it is going uphill or downhill

Table: Look to see if y-values

increase or decrease (as x-values

increase.)

Equation: Look for the sign on

the number in front of the ∛; positive is increasing, negative is decreasing.

14. Increasing, (7, –4)

15. Decreasing, (–9, 0)

16. Shifted Left 2, Down 4

17. Flipped; Shifted Right 3, Up 4

18. Graph Number 4 1 3 2

Point of Inflect. (-2,6) (-1,3) (5,0) (0,-4)

y-intercept (0,4.8) (0,4) (0, 1.8) (0,-4)

Inc/Decreasing Dec. Inc. Dec. Inc.

19. Answers will vary:

a) ( = √� + 3^ + 5

b) ( = −√� − 4^ − 6

c) ( = √� − 2^ + 9

-12 -8 -4 4

-8

-4

4

8

x

y

-8 -4 4 8

-8

-4

4

8

x

y

-8 -4 4 8

-8

-4

4

8

x

y

-12 -8 -4 4

-8

-4

4

8

x

y

-12 -8 -4 4

-8

-4

4

8

x

y



SECTION 7.2A

1. [A] 2. [C] 3. [A]

4. [B] 5. [A] 6. [C]

7. a) False: should = 2 ∙ 2 ∙ 2

b) True

c) False: should = �g(HF

d) True

e) False: should = 15�

f) False: should = MhijVik

8. Katiana

(�H)N = �W[�E

�N = �N

9. a) 4

b) –3

c) 0

d) any real number except 0

10. Sarah made an error when

simplifying the denominator in

Line 2: the –2 exponent

should have gone to the p and

d, not the q. The denominator

should simplify to lW0<G4WK

and a final answer of 4 11p q

11. Tyrell also made an error in

Line 2: 2WG should simplify to 0

0K (not –8). The final answer

should be 0M �O.


-Simplify: (�(H)W to �W(WK

So: 5 2 6

4 5

3

18

x x y

x y

− −

−

-Simplify x’s in numerator:

�E�W = �H So: 3 6

4 5

3

18

x y

x y

−

−

-Reduce numbers: H

0M = 0K

-Simplify x’s: V^

Vmn = �g

-Simplify y’s: ompok = (W00 = 0

oii

-Final Answer: Vq

Koii

13. Yes – Variable parts match so

they are “like terms” = 5�H(G

14. 4

15. 5

1

2k

16. 5

1

16 p

17. 9 3

12

c p

r

18. 6 129b z

19. 49

2

x

SECTION 7.2B

1. [H] 2. [B] 3. [D] 4. [E]

5. [G] 6. [A] 7. [F] 8. [C]

9. [C] 10. [B]

11. No – Exponent is NOT neg, so

exponent should not be moved

to denominator.

Ans = √17

12. Yes

13. Yes

14. No 5ik ≠ 1 . Ans = √5sk

15. Yes

16. Yes

17. No: 4 5 454 4=

18. [D] 19. [C] 20. ☺

21. (54)0/G or 50/G40/G

22. ( )1 6 1 6 1 6 1 636 36fg f g= ; ( )1 636 6≠

23. ☺

24. ( )1 5

2 2b c+ ; cannot be

simplified since there is no

power of a sum exponent

property.

25. ☺

26. 0H 27. 5 28.

00NG

29. Gg 30.

HGH0E 31. 7d

32. M

g|u^| 33. 0

0K n

34. 806.76≈ 806 people

35. 13,770mm

36. True: −360/ = −1 ∙ 360/

= −6

37. False: √−16n is not real

SECTION 7.2C

1. Never: √−256n is not real

2. Always

3. Always

4. Never: 3 343 7=

5. Sometimes

6. Sometime

7. Sometimes

#5 - #7: Since the radical on the left

side of the equation has an

even root, the answer must be

greater than or equal to zero.

However, the right side of the

equation has a variable to an

odd power, which could be

either a positive or a negative

answer.

8. Always

9. 3 10. 16 11. 9

12. 8 13. 25 14. 0.3

15. 0.6�G 16. 0NH 17. 6v4E/H

18. 64YGℎ



19. 343|4H| 20.

EG

21. KGxp

22. [A], [C], and [D] require

absolute value symbols

around one of the variables.

23. ≈ 1.34 in

24. ≈ 1.68 ft

25. ≈ 0.48 cm

26. 234 cm

27. 864 cm2

28. ≈ 0.165 yz 16.5%

SECTION 7.2D

ANSWER KEY

1. 2 2. 2.520

3. 0O ≈ 0.111 4.

2

1 1

49 2401=

5. 16 6. 1.5

7. √6 ≈ 2.449 8. 0N

KE ≈ 0.016

9. 100 10. 6b

Quiz grade: 2/10

Incorrect: (1, 3, 4, 6, 7, 8, 9, 10)

11. Neither Jerome nor Cambria

have correct answers:

( )( )

1 6

1 6

164

64

−− =

−

which is not a real number.

(−64)K = 68,719,476,736

12. ? = 0

13. ? = 0H

14. no number possible to replace

the question mark

15. ? = 0H

16. ? =− H

17. ? =− 0

18. 36

19. 4

20. 53 1212a

21. 0

\ji/ip

22. 19 6k

23. z

24. |s| 25. 2

26. �H/

27. 6

28. 9

29. 1

30. Over 162 trillion years

SECTION 7.3A

1. a) Intersection at (16, 4); x = 16

b) Intersections at (–2, 0) and

(–1, 1); x = –2 or x= –1

c) Intersections at (–3, 4) and

(13, 4); x = –3 or x=13

d) No intersections;

No Real Solutions

e) Intersection at (2,√2); x = 2

f) Intersection at (31.5, 5);

x = 31.5

2. a) x = 6 b) x = 36

c) x =0.807 or x =6.193

d) x = 2.5 e) x = 1

f) x = 11 or x = –16

g) x = 2 or x = 18

h) No Real Solution

i) x = –39

3. 4 ≈ 72.727 ft

4. | ≈ 149.34 million km

5. a) @ ≈ 261.2 km/hr

b) 4 ≈ 0.108 km

6. a) @ ≈ 1.924 sec

b) } ≈ 5.066 ft

7. No – If the graphing calculator

yields decimal solutions that do

not terminate or repeat, the

solutions (irrational) can be

found algebraically.

SECTION 7.3B

1. x = 9; Yes, one intersection on

graph at x = 9


graph at x = 5

3. x = 5; NO, graphs do not

intersect so NO REAL

SOLUTION

4. x = 5 or x = 2; Yes, two ints on

graph at x = 5 and x = 2

5. x = 7 or x = -3; No, only one

intersection at x = 7 so x = 3 is

extraneous.


graph at x = 2


graph at x = 10

8. x = 5 or x = –4; Yes, two ints on

graph at x = 5 and x = –4

9. For problem #3, when x = 5, the

‘check’ yields 8 = 2. This is a

false sentence therefore the

value of 5 is not a solution to

the equation.

For problem #5, when x = –3,

the ‘check’ yields –6 = 6. This

is a false sentence therefore

the value of –3 is not a




10. a) Graphically: The graphs do

not intersect for the

extraneous x-value.

b) Algebraically: A solution

comes out of the work but

doesn’t check so is not a

true solution.

11. Serge did the problem

correctly. Jeremy needed to

subtract 25 BEFORE squaring

both sides

12. Indy did the problem correctly.

Latishia did the work correctly

but did not check for

extraneous solutions; x = 3 is

extraneous. Sango squared

(2 – x) incorrectly;

(2 – x)2 = (2 – x) (2 – x)

not 22 – x2

13. x = 5

14. n = 77

15. No real solution


17. x = 7 18. x = –13

19. x = –8 20. x = 1 21. r = 8


23. x = H

E

24. x = ±32

25. d = 2 ± 2√2

26. b = 20

27. x = 621

28 – 33: Answers may vary

28. graphically

29. algebraically

30. graphically

31. algebraically

32. algebraically

33. algebraically


√5� + 3 = −2

35. ℎ ≈ 0.6;

36. a) f = 30 gallons/min

b) p = 640,000 lbs/in2

37. 4.15t ≈ ; It would take

4.15 hours.

38. x = 58.81 mi2

Unit 7 - Review

1. a)

Domain: x > 0

Range: y > 0

b)

Domain: x > –1

Range: y > 0

c)

Domain: x > 0

Range: y > –2

d)

Domain: x > –2

Range: y > –3

e)

Domain: x > 0

Range: y < 0

f)

Domain: x > –1

Range: y < 2

-2 2 4 6 8 10

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

x

y

-2 2 4 6 8 10

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

x

y

-2 2 4 6 8 10

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

x

y

-2 2 4 6 8 10

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

x

y

-2 2 4 6 8 10

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

x

y

-2 2 4 6 8 10

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

x

y



2. Answers may vary:

( � √� � 2 + 6

3. a)

Domain: 7

Range: 7

Inflection Point: (0, 0)

b)

Domain: 7

Range: 7

Inflection Point: (–1, 0)

c)

Domain: 7

Range: 7


d)

Domain: 7

Range: 7


e)

Domain: 7

Range: 7


f)

Domain: 7

Range: 7


4. 0

GO

5. 0

0EV^

6. ENVk

7. 343

8. 5i

9. √�0N^

10. G~kE�p

11. 27�0E�g

12. 0

g

13. 6�G

14. Vhn/^

oj

15. 16�E

16.M

gV^

17. a

18. −0

E

19. 38vG4 ft

20. 104,573 people

21. � =M

H

22. 1 or 2x x= − = −

23. x = 7

24. x = 5 or x = –1

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

-5

-4

-3

-2

-1

1

2

3

4

5

x

y

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

-5

-4

-3

-2

-1

1

2

3

4

5

x

y

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

-5

-4

-3

-2

-1

1

2

3

4

5

x

y

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

-5

-4

-3

-2

-1

1

2

3

4

5

x

y

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

-5

-4

-3

-2

-1

1

2

3

4

5

x

y

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

-5

-4

-3

-2

-1

1

2

3

4

5

x

y



SECTION 8.1A

1. Opens: Up

Axis: x = –2

Vertex: (–2, –6)

2. Opens: Down

Axis: x = 3

Vertex: (3, 4)

3. Opens: Up

Axis: x = 5

Vertex: (5, –4)

4. Opens: Down

Axis: x = 2

Vertex: (2, –5)

5. Opens: Down

Axis: x = 0

H

Vertex: (0

H, –4)

6. Opens: Up

Axis: x = 2

Vertex: (2, 4)

7.

Domain: x = ℝ

Range: y > –4

8.

Domain: x = ℝ

Range: y > 0

9.

Domain: x = ℝ

Range: y > 4

10.

Domain: x = ℝ

Range: y < –3

11.

Domain: x = ℝ

Range: y < 3

12.

Domain: x = ℝ

Range: y < 5

13. Domain: x = ℝ

Range : y > 3

14.

� Vertex: (2, 0)

� Opens: Up

� Domain: x = ℝ

� Range: y > 0

� Steepness: m = ±3

15.

� Vertex: (6, 0)

� Opens: Up

� Domain: x = ℝ

� Range: y > 0

� Steepness: m = ±3

16. Both have a –6 in the absolute

value bars with the x.

17. #14 has the multiplier of 3

inside the absolute values bars;

#15 has the multiplier outside

the absolute value bars.

18. When the 3 is on the inside, it

influences the x-coordinate of

the vertex by dividing the 6 by a

factor of 3.

19. [equation B]

20. [D]

21. y = 2x + 8, x > –4

y = –2x – 8, x < –4

-8 -4 4 80-8

-4

4

8

0x

y

-8 -4 4 80-8

-4

4

8

0x

y

-8 -4 4 80-8

-4

4

8

0x

y

-8 -4 4 80-8

-4

4

8

0x

y

-8 -4 4 80-8

-4

4

8

0x

y

-8 -4 4 80-8

-4

4

8

0x

y



SECTION 8.1B

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.



13.

Domain: x = ℝ

Range: y = ℝ

14.

Domain: x = ℝ

Range: 0y >

15. [B] and [D]

16. Answers may vary.

a. ( / |� � 2| � 5

b. ( > |� − 4| � 3

c. ( / |� � 2| � 6

17. [C]

18. [B]

19. [A]

20. [D]

SECTION 8.2A

1. Graphical Solution

Algebraic Solution

x = 3 x = –3


Algebraic Solution

x = 2 x = –2


Algebraic Solution

x = 2 x = –8


Algebraic Solution

x = g

H x = –1


Algebraic Solution

–5 < x < –1


Algebraic Solution

x < 2 or x > 6

-4 -2 2 40-2

2

4

0x

y

-4 -2 2 40-2

2

4

6

8

10

0x

y

-10 -8 -6 -4 -2 2 40-2

2

4

6

8

0x

y

-4 -2 2 40-2

2

4

6

8

10

0x

y

-10 -8 -6 -4 -2 2 4 0 -2

2

4

6

8

0

x

y

-2 2 4 6 8 0-2

2

4

6

8

10

0

x

y




Algebraic Solution

x < 6 or x > 10


Algebraic Solution

–1 < x < 0


Algebraic Solution

–2 < x < G

H

10. –10 < x < –4

11. x > 9 or x < –1

12. W00

K/ � /

g

K

13. W

H < x < 4

14. –4 < x < 8

15. x < –6 or x > 10

16. {all REAL numbers}

17. 2 < x < 6


16, 16.1, 16.6, 15.9, 15.4

b)

c) 15.4 < x < 16.6

d)|� − 16| ≤ 0.6


40℉, 41℉, 42℉, 74℉, 75℉

b)

c) x < 43 or x > 73

d) |� − 58| > 15


3, 4, 4.5, 5, 6

b)

c) 3 < x < 6

d) |� − 4.5| ≤ 1.5

-2 2 4 6 8 10 120-2

2

4

6

0x

y

-4 -2 2 4 0-2

2

4

6

8

0x

y

-4 -2 2 4 0-6

-4

-2

2

4

6

0

x

y



Unit 8 - Review

1. a) Up

b) (3, –1)

c) x = 3

d) ±2

2. a) Down

b) (–1, 4)

c) x = –1

d) ±0

H

3.

4.


( = 4|� − 1| − 3


( =−1

3|� + 2| + 1

7. x = N

H x =

W0N

H

8. x = 8 x = 0

9. x > 2 or x < –7

10. –9 < x < 3

11. a) 230 < x < 270

b) |� − 250| ≤ 20

12. a) 7.7 < x < 8.3

b) |� − 8| < 0.3

-2 2 4 6 8-2

2

4

6

8

x

y

-6 -4 -2 2 4 6-2

2

4

6

8

x

y

-6 -4 -2 2 4 6-6

-4

-2

2

4

6

x

y

Intermediate Algebra (B) Unit 5 – Selected Answers · 2015-01-08 · Intermediate Algebra (B)...

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Transcript of Intermediate Algebra (B) Unit 5 – Selected Answers · 2015-01-08 · Intermediate Algebra (B)...