Alg2 - CH8 Practice Test
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. Given: y varies directly as x, and y = −5 when x = 2.5. Write and graph the direct variation
function.
a. y = –2x
b. xy = −12.5
c. y = –0.5x
____
____
____
____
____
d. Not enough information. It takes two points to determine a line.
2. The number of lawns l that a volunteer can mow in a day varies inversely with the number of
shrubs s that need to be pruned that day. If the volunteer can prune 6 shrubs and mow 8 lawns in
one day, then how many lawns can be mowed if there are only 3 shrubs to be pruned?
a. 4 lawns
c. 11 lawns
b. 5 lawns
d. 16 lawns
8x4y2 9xy2z6
⋅
3. Multiply
. Assume that all expressions are defined.
3z3
4y4
a. 6x4yz 2
c. 6x5z 3
b. 6x5y8z 9
d. 3 x3y2z
2
2
4. The area of a rectangle is equal to x + 10x + 16 square units. If the length of the rectangle is equal
to x + 8 units, what expression represents its width?
a. x − 2
c. x + 4.00
b. x + 2
d. x + 8
− 59 .
5. Add x + 6 + −12x
2
x − 7 x − 3x − 28
a. −11x − 53
c.
x+6
2
(x − 7)(x + 4)
x − 2x − 35
b. x2 + 10x + 24
d. x + 5
x+4
(x + 4)(x − 7)
−5 + x − 6
x − 4 10
6. Simplify
. Assume that all expressions are defined.
x+3
x−4
a.
x − 56
10( x + 3 )
c. x2 − 10x − 26
10( x + 3 )
b.
____
d.
x2 − 10x − 26
10( x2 − x − 12 )
1 + 3.
x+7
c. Vertical asymptote: x = 7
Domain: {x| x ≠ 7 }
Horizontal asymptote: y = −3
7. Identify the asymptotes, domain, and range of the function g(x) =
a. Vertical asymptote: x = −7
Domain: {x| x ≠ −7 }
Horizontal asymptote: y = −3
Range: {y| y ≠ −3 }
b. Vertical asymptote: x = 7
Domain: {x| x ≠ 7 }
Horizontal asymptote: y = 3
Range: {y| y ≠ 3 }
____
x3 − 14x2 + 14x + 104
10( x2 − x − 12 )
8. Identify holes in the graph of f( x ) =
Range: {y| y ≠ −3 }
d. Vertical asymptote: x = −7
Domain: {x| x ≠ −7 }
Horizontal asymptote: y = 3
Range: {y| y ≠ 3 }
x2 + 8x + 12 . Then graph.
x+2
a. There is a hole in the graph at x = −6.
b. There is a hole in the graph at x = −2.
c. There are no holes in the graph.
d. There is a hole in the graph at x = 2.
____
9. Solve the equation x − 9 = − 18 .
x
a. x = 6
b. x = 3 or x = 6
____
____
c. x = −3 or x = −6
d. x = 3
10. Jeremy and Ahmed are painting a wall. Working alone, Jeremy can paint 14 of the wall in 1 hour.
Working together, Jeremy and Ahmed can paint the entire wall in 35 minutes. How long would it
take Ahmed to paint the entire the wall by himself?
a. 20 minutes
c. 3 hours
b. 35 minutes
d. 4 hours
2
3
11. The surface area S of a cube with volume V is S = 6V . What effect does increasing the volume of
a cube by a factor of 7 have on the on the surface area?
2
a.
3
The surface area increases by a factor of 7 .
b. The surface area increases by a factor of 7.
c. The surface area increases by a factor of 72.
d.
−1
3
The surface area increases by a factor of 7 .
____
12. The function g is a translation 3 units left and 2 units up of f(x) =
a. g(x) = x + 12 + 2
c. g(x) = x + 14
b. g(x) =
d. g(x) =
x + 12 − 2
x + 9 . Write the function g(x).
x+8
Numeric Response
13. The cost per student of a camping trip varies inversely as the number of students who attend. It
will cost each student $85 if 28 students attend. How many students would have to attend to get the
cost down to $74.38?
14. A camping group hikes the same distance on each day of a four-day hiking trip. The table shows
the group’s average speed on each day. What is the group’s average speed for the entire trip,
rounded to the nearest tenth of a kilometer per hour?
Hiking Results
Day
Speed (km/h)
1
6.2
2
5.9
3
4.4
4
3.9
15. How many vertical asymptotes does the graph of f(x) =
x2 + 9 have?
x2 − 4x + 4
1
16. What value of x makes ( 2x + 17 ) 3 = 3 a true statement?
Matching
Match each vocabulary term with its definition.
a. constant of variation
b. direct variation
c. inverse variation
d. joint variation
e. combined variation
f. positive variation
g. negative variation
____
____
____
____
17. a relationship between two variables, x and y, that can be written in the form y = kx, where k is a
nonzero constant
18. the constant k in a direct variation equation
19. a relationship containing both direct and inverse variation
20. a relationship among three variables that can be written in the form y = kxz, where k is a nonzero
constant
Match each vocabulary term with its definition.
a. rational function
b. quadratic function
c. continuous function
d. hole (in a graph)
e. square-root function
f. complex fraction
g. discontinuous function
____
____
____
____
21.
22.
23.
24.
a function whose graph has one or more jumps, breaks, or holes
a fraction that contains one or more fractions in the numerator, the denominator, or both
an omitted point on a graph
a function whose rule contains a variable under a square-root sign
Match each vocabulary term with its definition.
a. rational equation
b. rational inequality
c. radical equation
d. radical function
e. radical inequality
f. index
g. extraneous solution
____
____
____
____
25.
26.
27.
28.
n
n in the radical x , which represents the nth root of x
an inequality that contains a variable within a radical
an equation that contains a variable within a radical
a function whose rule contains a variable within a radical
Alg2 - CH8 Practice Test
Answer Section
MULTIPLE CHOICE
1. ANS: A
y = kx
−5 = k( 2.5 )
−2 = k
y = −2x
y varies directly as x.
Substitute –5 for y and 2.5 for x.
Solve for the constant of variation k.
Substitute –2 for k in the equation for direct variation.
Feedback
A
B
C
D
Correct!
Substitute the given values of x and y in the equation for direct variation, y = kx.
Then solve for k.
Substitute the given values of x and y in the equation for direct variation, y = kx.
Then solve for k.
Substitute the given values of x and y in the equation for direct variation, y = kx.
Then solve for k.
PTS: 1
DIF: Average
REF: Page 569
OBJ: 8-1.1 Writing and Graphing Direct Variation
TOP: 8-1 Variation Functions
2. ANS: D
One method is to use s 1l1 = s 2l2.
Substitute given values.
(6)(8) = (3)l2
Simplify.
48 = 3l2
Divide.
16 = l2
NAT: 12.5.1.e
Feedback
A
B
C
D
With inverse variation, when one quantity decreases, the other quantity
increases. The number of shrubs was divided by 2, so the number of lawns
should be multiplied by 2.
With inverse variation, when one quantity decreases, the other quantity
increases. The number of shrubs was divided by 2, so the number of lawns
should be multiplied by 2.
The number of shrubs was divided by 2, so the number of lawns should be
multiplied by 2.
Correct!
PTS: 1
NAT: 12.5.1.e
DIF: Average
REF: Page 571
TOP: 8-1 Variation Functions
OBJ: 8-1.5 Application
3. ANS: C
8 ⋅ 9(x4 ⋅ x)(y2 ⋅ y2)z6
Arrange the expressions so like terms are together:
.
3 ⋅ 4 ⋅ z 3 y4
Multiply the numerators and denominators, remembering to add exponents when multiplying:
72x5y4z6
.
12z3y4
Divide, remembering to subtract exponents: 6x5y0z 3.
Since y0
= 1, this expression simplifies to 6x5z3.
Feedback
A
B
C
D
A variable raised to the 0 power simplifies to 1.
When dividing powers with the same base, subtract the exponents.
Correct!
Multiply, then simplify.
PTS:
OBJ:
TOP:
4. ANS:
1
DIF: Basic
REF: Page 578
8-2.3 Multiplying Rational Expressions
NAT: 12.5.3.c
8-2 Multiplying and Dividing Rational Expressions
B
Area of a rectangle equals length times width.
A = lw
2
Substitute the area and length expressions given.
(x + 10x + 16) = (x + 8)w
Factor and solve for w.
(x + 8)(x + 2)
(x + 8)
(x + 2) = w
=w
Simplify.
Feedback
A
B
C
D
Multiply the expressions for length and width to be sure they create the
polynomial expression for area.
Correct!
Factor the polynomial to find the possible length and width.
In a rectangle, length times width equals area.
PTS: 1
DIF: Advanced
NAT: 12.5.3.c
TOP: 8-2 Multiplying and Dividing Rational Expressions
5. ANS: D
x + 6 + −12x − 59
x − 7 (x + 4)(x − 7)
Ê
ˆ
= ÁÁÁ x + 4 ˜˜˜ x + 6 + −12x − 59
Ë x + 4 ¯ x − 7 (x + 4)(x − 7)
Factor the denominators. The LCD is (x + 4)(x − 7).
Ê
Multiply by ÁÁÁ x + 4
ˆ˜
˜˜.
Ë x+4 ¯
x2 + 10x + 24 + −12x − 59
(x + 4)(x − 7) (x + 4)(x − 7)
x2 − 2x − 35
=
(x + 4)(x − 7)
=
(x + 5)(x − 7)
(x + 4)(x − 7)
= x+5
x+4
=
Add the numerators.
Factor the numerator.
Divide the common factor.
Feedback
A
B
C
D
Find a common denominator before adding the fractions.
This is the first fraction rewritten with the common denominator. Add this to the
second fraction.
Find a common denominator before adding the fractions.
Correct!
PTS: 1
DIF: Average
REF: Page 584
OBJ: 8-3.3 Adding Rational Expressions NAT: 12.5.3.c
TOP: 8-3 Adding and Subtracting Rational Expressions
6. ANS: C
Method 1 Write the complex fraction as division.
ÊÁ −5 x − 6 ˆ˜ x + 3
ÁÁ
˜˜ ÷
+
Ë x − 4 10 ¯ x − 4
Ê
ˆ
= ÁÁÁ −5 + x − 6 ˜˜˜ ⋅ x − 4
Ë x − 4 10 ¯ x + 3
ÊÁ
Ê ˆ
Ê
ˆ ˆ˜
= ÁÁÁ −5 ÁÁÁ 10 ˜˜˜ + x − 6 ÁÁÁ x − 4 ˜˜˜ ˜˜˜ ⋅ x − 4
Á x − 4 Ë 10 ¯ 10 Ë x − 4 ¯ ˜ x + 3
Ë
¯
2
x − 10x − 26 ⋅ x − 4
=
x+3
10( x − 4 )
2
x − 10x − 26 or x2 − 10x − 26
=
10x + 30
10( x + 3 )
Divide.
Multiply by the reciprocal.
Add by finding the LCD.
The common factor (x − 4) cancels.
Simplify.
Method 2 Multiply the numerator and denominator of the complex fraction by the LCD of the
fractions in the numerator and denominator.
−5 ( 10 ) ( x − 4 ) + x − 6 ( 10 ) ( x − 4 )
x−4
10
x + 3 ( 10 ) ( x − 4 )
x−4
=
−5( 10 ) + ( x − 6 ) ( x − 4 )
10( x + 3 )
The LCD is 10( x − 4 ).
Cancel common factors.
=
x2 − 10x − 26 or x2 − 10x − 26
10x + 30
10( x + 3 )
Simplify.
Feedback
A
B
C
D
Multiply both addends in the numerator by the reciprocal of the denominator,
not just the first.
Multiply the numerator by the reciprocal of the denominator.
Correct!
Multiply the numerator by the reciprocal of the denominator.
PTS:
OBJ:
TOP:
7. ANS:
1
DIF: Average
REF: Page 586
8-3.5 Simplifying Complex Fractions
8-3 Adding and Subtracting Rational Expressions
D
Write the function in the form g(x) =
NAT: 12.5.3.c
1 + k where x = h is the vertical asymptote and helps find
x−h
the domain, and y = k is the horizontal asymptote and helps find the range.
g(x) =
1 + 3, so h = −7 and k = 3.
x − (7)
Vertical asymptote: x = −7
Domain: {x| x ≠ −7 }
Horizontal asymptote: y = 3
Range: {y| y ≠ 3 }
Feedback
A
B
C
D
The horizontal asymptote is equal to the vertical translation of the parent
function.
The vertical asymptote is at the value of x that makes the denominator equal 0.
The vertical asymptote is at the value of x that makes the denominator equal 0.
The horizontal asymptote is equal to the vertical translation of the parent
function.
Correct!
PTS: 1
DIF: Average
REF: Page 593
OBJ: 8-4.2 Determining Properties of Hyperbolas
8. ANS: B
2
f( x ) = x + 8x + 12
x+2
( x + 2 )( x + 6 )
x+2
= x+6
=
TOP: 8-4 Rational Functions
Factor the numerator. x + 2 is a factor in both the numerator
and the denominator, so there is a hole at x = −2.
Divide out common factors.
Except for the hole at x = −2, the graph of f is the same as y = x + 6. On the graph, indicate the hole
ÔÏ
Ô¸
with an open circle. The domain of f is ÔÌ x | x ≠ −2 Ô˝.
Ó
˛
Feedback
A
B
C
D
For what value(s) of x are the numerator and denominator of f(x) equal to zero?
Correct!
For what value(s) of x are the numerator and denominator of f(x) equal to zero?
For what value(s) of x are the numerator and denominator of f(x) equal to zero?
PTS: 1
DIF: Average
REF: Page 596
OBJ: 8-4.5 Graphing Rational Functions with Holes
9. ANS: B
x( x ) − 9( x ) = − 18 ( x )
x
TOP: 8-4 Rational Functions
Multiply each term by the LCD.
Simplify. Note x ≠ 0
Write in standard form.
Factor.
Apply the Zero-Product Property.
Solve for x.
x2 − 9x = −18
x2 − 9x + 18 = 0
( x − 3 )( x − 6 ) = 0
x − 3 = 0 or x − 6 = 0
x = 3 or x = 6
Check:
x − 9 = − 18
x
3−9
−6
x − 9 = − 18
x
− 18
3
−6
6−9
−3
− 18
6
−3
Feedback
A
B
C
D
Multiply each term by the LCD. Then solve the resulting quadratic equation.
There may be more than one solution.
Correct!
Multiply each term by the LCD. Then solve the resulting quadratic equation.
Multiply each term by the LCD. Then solve the resulting quadratic equation.
There may be more than one solution.
PTS: 1
DIF: Average
REF: Page 600
OBJ: 8-5.1 Solving Rational Equations NAT: 12.5.4.a
TOP: 8-5 Solving Rational Equations and Inequalities
10. ANS: C
Jason’s rate: 14 of the wall per hour
Let Ahmed’s rate be
1
h
of the wall per hour.
Given: Jason’s rate ( 14 ) plus Ahmed’s rate ( 1h ) = the entire wall in 35 minutes.
1
4
+ 1h = 127
12h ⋅ ( 14 + 1h ) = 12h ⋅
7
12
Multiply by the LCD.
Simplify.
3h + 12 = 7h
Solve for h.
h=3
Ahmed can paint the entire wall by himself in 3 hours.
Feedback
A
B
C
D
Solve for h, not 1/h.
That's how long it takes them to paint the entire wall working together.
Correct!
That's how long it takes Jeremy to paint the entire wall working alone.
PTS: 1
DIF: Average
REF: Page 603
OBJ: 8-5.4 Application
NAT: 12.5.4.a
TOP: 8-5 Solving Rational Equations and Inequalities
KEY: algebra | equation | rate
11. ANS: A
To find the effect, replace V with 7V in the equation.
2
Substitute.
S = 6( 7V ) 3
2
3
S = 6(7) V
2
3
Use the Power of a Product Property.
2
3
The surface area changes by a factor of 7 when the volume changes by a factor of 7.
Feedback
A
B
C
D
Correct!
Replace V with the new volume, and then use the Power of a Product Property.
Replace V with the new volume, and then use the Power of a Product Property.
Replace V with the new volume, and then use the Power of a Product Property.
PTS: 1
DIF: Advanced
TOP: 8-6 Radical Expressions and Rational Exponents
12. ANS: A
Replace f(x) with f(x – h) and simplify.
x + 9 ⇒ (x − 3) + 9 = x + 12
x + 12 ⇒
x + 12 + 2
Replace f(x – h) with f(x – h) + k.
Feedback
A
B
C
D
Correct!
For the vertical translation, replace f(x) with f(x) + k.
Don't combine the vertical and horizontal units of translation.
For the horizontal translation, replace f(x) with f(x – h).
PTS:
1
DIF:
Advanced
TOP: 8-7 Radical Functions
DIF:
Average
NAT: 12.5.4.a
NUMERIC RESPONSE
13. ANS: 32
PTS: 1
14. ANS: 4.9
TOP: 8-1 Variation Functions
PTS: 1
DIF: Average
TOP: 8-3 Adding and Subtracting Rational Expressions
15. ANS: 1
PTS: 1
16. ANS: 5
DIF:
Advanced
TOP: 8-4 Rational Functions
PTS:
DIF:
Advanced
TOP: 8-8 Solving Radical Equations and Inequalities
1
MATCHING
17. ANS:
TOP:
18. ANS:
TOP:
19. ANS:
TOP:
20. ANS:
TOP:
B
PTS: 1
8-1 Variation Functions
A
PTS: 1
8-1 Variation Functions
E
PTS: 1
8-1 Variation Functions
D
PTS: 1
8-1 Variation Functions
DIF:
Basic
REF: Page 569
DIF:
Basic
REF: Page 569
DIF:
Basic
REF: Page 572
DIF:
Basic
REF: Page 570
21. ANS:
TOP:
22. ANS:
TOP:
23. ANS:
TOP:
24. ANS:
TOP:
G
PTS: 1
DIF: Basic
8-4 Rational Functions
F
PTS: 1
DIF: Basic
8-3 Adding and Subtracting Rational Expressions
D
PTS: 1
DIF: Basic
8-4 Rational Functions
E
PTS: 1
DIF: Basic
8-7 Radical Functions
REF: Page 593
25. ANS:
TOP:
26. ANS:
TOP:
27. ANS:
F
PTS: 1
DIF: Basic
8-6 Radical Expressions and Rational Exponents
E
PTS: 1
DIF: Basic
8-8 Solving Radical Equations and Inequalities
C
PTS: 1
DIF: Basic
REF: Page 610
REF: Page 586
REF: Page 596
REF: Page 619
REF: Page 630
REF: Page 628
TOP: 8-8 Solving Radical Equations and Inequalities
28. ANS: D
PTS: 1
DIF: Basic
TOP: 8-7 Radical Functions
REF: Page 619