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WGU C957 PRE-ASSESSMENT: APPLIED
ALGEBRA (FXO1) (PFXO) 2026 with complete
solution
1. A company uses the function S(t)S(t) to represent the number of
stores it expects to have after tt years. What does the statement
S(6)=42S(6)=42 mean?
A. The company opens 6 stores every 42 years.
B. After 42 years, the company will have 6 stores.
C. After 6 years, the company is expected to have 42 stores.
D. The company will open 42 stores every 6 years.
Answer: C
Solution: In function notation, the input is t=6t=6, and the output is
S(6)=42S(6)=42. Therefore, after 6 years, the model predicts 42 stores.
2. A fundraising organization models its monthly donations with
D(m)=125m+500D(m)=125m+500, where mm represents the number of
months. What does the coefficient 125 represent?
A. The initial amount of donations
B. The monthly rate at which donations increase
C. The total donations after 125 months
D. The number of months required to raise $500
Answer: B
Solution: In a linear function y=mx+by=mx+b, mm is the slope or rate
of change. Therefore, 125 represents an increase of $125 per month.
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3. A taxi company charges a $4 starting fee plus $2.50 for every mile
traveled. Which function represents the total cost C(x)C(x) for xx miles?
A. C(x)=4x+2.50C(x)=4x+2.50
B. C(x)=2.50x+4C(x)=2.50x+4
C. C(x)=6.50xC(x)=6.50x
D. C(x)=4−2.50xC(x)=4-2.50x
Answer: B
Solution: The fixed starting fee is the y-intercept, $4. The $2.50 per
mile is the slope. Therefore,
C(x)=2.50x+4C(x)=2.50x+4
4. A linear model passes through the points (2,11)(2,11) and
(6,27)(6,27). What is the rate of change?
A. 2
B. 3
C. 4
D. 5
Answer: C
Solution:
m=27−116−2m=\frac{27-11}{6-2} m=164=4m=\frac{16}{4}=4
The rate of change is 4 units per unit increase in xx.
5. A population increases from 8,000 to 10,500 over a period of 5 years.
What is the average rate of change?
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A. 250 people per year
B. 400 people per year
C. 500 people per year
D. 2,500 people per year
Answer: A
Solution:
10,500−8,0005=2,5005=500\frac{10,500-8,000}{5} =\frac{2,500}{5}
=500
Therefore, the average rate of change is 500 people per year.
6. A function is represented by f(x)=3x2−5x+2f(x)=3x^2-5x+2. What
type of function is this?
A. Linear
B. Polynomial quadratic
C. Exponential
D. Logistic
Answer: B
Solution: The highest exponent of xx is 2, so the function is a quadratic
polynomial.
7. For the function
f(x)=x2−8x+12,f(x)=x^2-8x+12,
which value of xx gives the minimum value?
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A. 2
B. 4
C. 6
D. 8
Answer: B
Solution: For ax2+bx+cax^2+bx+c, the vertex occurs at
x=−b2a.x=\frac{-b}{2a}.
Here, a=1a=1 and b=−8b=-8:
x=−(−8)2(1)=4.x=\frac{-(-8)}{2(1)}=4.
Because a>0a>0, the parabola opens upward, so x=4x=4 produces the
minimum.
8. What is the minimum value of
f(x)=x2−8x+12?f(x)=x^2-8x+12?
A. -4
B. 0
C. 4
D. 12
Answer: A
Solution:
Substitute x=4x=4:
f(4)=42−8(4)+12f(4)=4^2-8(4)+12 =16−32+12=−4.=16-32+12=-4.
Therefore, the minimum value is -4.
WGU C957 PRE-ASSESSMENT: APPLIED
ALGEBRA (FXO1) (PFXO) 2026 with complete
solution
1. A company uses the function S(t)S(t) to represent the number of
stores it expects to have after tt years. What does the statement
S(6)=42S(6)=42 mean?
A. The company opens 6 stores every 42 years.
B. After 42 years, the company will have 6 stores.
C. After 6 years, the company is expected to have 42 stores.
D. The company will open 42 stores every 6 years.
Answer: C
Solution: In function notation, the input is t=6t=6, and the output is
S(6)=42S(6)=42. Therefore, after 6 years, the model predicts 42 stores.
2. A fundraising organization models its monthly donations with
D(m)=125m+500D(m)=125m+500, where mm represents the number of
months. What does the coefficient 125 represent?
A. The initial amount of donations
B. The monthly rate at which donations increase
C. The total donations after 125 months
D. The number of months required to raise $500
Answer: B
Solution: In a linear function y=mx+by=mx+b, mm is the slope or rate
of change. Therefore, 125 represents an increase of $125 per month.
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3. A taxi company charges a $4 starting fee plus $2.50 for every mile
traveled. Which function represents the total cost C(x)C(x) for xx miles?
A. C(x)=4x+2.50C(x)=4x+2.50
B. C(x)=2.50x+4C(x)=2.50x+4
C. C(x)=6.50xC(x)=6.50x
D. C(x)=4−2.50xC(x)=4-2.50x
Answer: B
Solution: The fixed starting fee is the y-intercept, $4. The $2.50 per
mile is the slope. Therefore,
C(x)=2.50x+4C(x)=2.50x+4
4. A linear model passes through the points (2,11)(2,11) and
(6,27)(6,27). What is the rate of change?
A. 2
B. 3
C. 4
D. 5
Answer: C
Solution:
m=27−116−2m=\frac{27-11}{6-2} m=164=4m=\frac{16}{4}=4
The rate of change is 4 units per unit increase in xx.
5. A population increases from 8,000 to 10,500 over a period of 5 years.
What is the average rate of change?
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A. 250 people per year
B. 400 people per year
C. 500 people per year
D. 2,500 people per year
Answer: A
Solution:
10,500−8,0005=2,5005=500\frac{10,500-8,000}{5} =\frac{2,500}{5}
=500
Therefore, the average rate of change is 500 people per year.
6. A function is represented by f(x)=3x2−5x+2f(x)=3x^2-5x+2. What
type of function is this?
A. Linear
B. Polynomial quadratic
C. Exponential
D. Logistic
Answer: B
Solution: The highest exponent of xx is 2, so the function is a quadratic
polynomial.
7. For the function
f(x)=x2−8x+12,f(x)=x^2-8x+12,
which value of xx gives the minimum value?
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A. 2
B. 4
C. 6
D. 8
Answer: B
Solution: For ax2+bx+cax^2+bx+c, the vertex occurs at
x=−b2a.x=\frac{-b}{2a}.
Here, a=1a=1 and b=−8b=-8:
x=−(−8)2(1)=4.x=\frac{-(-8)}{2(1)}=4.
Because a>0a>0, the parabola opens upward, so x=4x=4 produces the
minimum.
8. What is the minimum value of
f(x)=x2−8x+12?f(x)=x^2-8x+12?
A. -4
B. 0
C. 4
D. 12
Answer: A
Solution:
Substitute x=4x=4:
f(4)=42−8(4)+12f(4)=4^2-8(4)+12 =16−32+12=−4.=16-32+12=-4.
Therefore, the minimum value is -4.
