AP Precalculus Unit 2 practice questions

Question 1

The function $$f(x) = \log_2(x)$$ has domain $$(0, \infty)$$ and a vertical asymptote at $$x = 0$$. If $$r(x) = \log_2(4 - x)$$, which statement correctly describes the domain of $$r$$ and explains why it differs from $$f$$?

Accepted Answer

Domain $$(-\infty, 4)$$ because the argument $$4 - x$$ must be positive, requiring $$4 - x > 0$$, which means $$x < 4$$.

Answer

Domain $$(0, \infty)$$ because logarithmic functions always require positive arguments, and the transformation does not change this requirement.

Answer

Domain $$(4, \infty)$$ because the $$4 - x$$ inside the logarithm shifts the domain right by 4 units from the original domain $$(0, \infty)$$.

Answer

Domain $$(-\infty, 4)$$ because $$4 - x > 0$$; vertical asymptote at $$x = 4$$ because the asymptote shifts from $$x = 0$$ to $$x = 4$$.

Question 2

The function $$f(x) = \log_2(x)$$ has domain $$(0, \infty)$$ and vertical asymptote at $$x = 0$$. If $$t(x) = 2f(\frac{x}{4} - 1) - 3$$, what is the domain of $$t$$ and where is its vertical asymptote?

Accepted Answer

Domain: $$(4, \infty)$$; Vertical asymptote: $$x = 4$$

Answer

Domain: $$(0, \infty)$$; Vertical asymptote: $$x = 0$$

Answer

Domain: $$(1, \infty)$$; Vertical asymptote: $$x = 1$$

Answer

Domain: $$(4, \infty)$$; Vertical asymptote: $$x = 1$$

Question 3

Let $$f(x) = 3\sin(x)$$ and $$g(x) = x + \frac{\pi}{4}$$. Which expression represents $$(f \circ g)(x)$$?

Accepted Answer

$$3\sin\left(x + \frac{\pi}{4}\right)$$

Answer

$$3\sin(x) + \frac{\pi}{4}$$

Answer

$$3\sin(x) + 3\frac{\pi}{4}$$

Answer

$$\sin\left(3x + \frac{\pi}{4}\right)$$

Question 4

A population model uses $$P(t) = 1000 \cdot 2^t$$. A biologist discovers that a second population follows the same exponential pattern but starts at 2500 individuals and grows at half the rate. Which transformed function models the second population?

Accepted Answer

$$Q(t) = 2500 \cdot 2^{t/2}$$

Answer

$$Q(t) = 2500 \cdot 2^{2t}$$

Answer

$$Q(t) = 1000 \cdot 2^t + 1500$$

Answer

$$Q(t) = 2.5 \cdot 2^t$$

Question 5

A population of bacteria is modeled by $N(t) = 1000 \cdot 2^t$ , where t is days. A researcher applies a treatment that eliminates half the bacteria immediately and then slows growth to one-third the original rate. Which function models the treated population?

$N(t) = 500 \cdot 2^{t/3}$

$N(t) = 1000 \cdot 2^{t - 1}$

$N(t) = 500 \cdot 2^t$

$N(t) = 1000 \cdot 2^{0.33t}$

Accepted Answer

$$N(t) = 500 \cdot 2^{t/3}$$

Answer

$$N(t) = 1000 \cdot 2^{t - 1}$$

Answer

$$N(t) = 500 \cdot 2^t$$

Answer

$$N(t) = 1000 \cdot 2^{0.33t}$$

Question 6

A financial analyst models investment growth using $$V(t) = 1000 \cdot 1.05^t$$, where $$t$$ is years and $$V(t)$$ is value in dollars. To model a different investment with a 2-year delayed start, the analyst uses $$W(t) = 1000 \cdot 1.05^{t-2}$$. Which assumption about this transformation is essential for the model to accurately represent the delayed investment?

Accepted Answer

The 5% annual growth rate remains constant; only the start time shifts.

Answer

The initial investment is reduced by 2 dollars from the principal.

Answer

The growth rate is reduced to 3% annually instead of 5%.

Answer

The investment value is doubled at all times due to the delay.

Question 7

A financial analyst models the value of an investment as $$V(t) = 1500(1.06)^{t-2} + 400$$, where $$V$$ is value in dollars and $$t$$ is years since 2020. This model is derived from the parent function $$f(t) = (1.06)^t$$ through additive and multiplicative transformations. Which assumption about the investment's behavior is critical for this model to remain valid beyond the observed data?

Accepted Answer

Assumes the growth rate remains constant at 6% annually and that external conditions (market, interest rates) do not change; if these change, extrapolations will be unreliable.

Answer

Assumes the growth rate of 6% annually is an average that may fluctuate year-to-year, but these fluctuations average out over time; if growth becomes consistently higher or lower, the model remains valid because averaging still occurs.

Answer

Assumes the growth rate remains constant at 6% annually and that external conditions do not change; if these conditions do change, the model parameters (1500 and 400) must be recalibrated but the exponential form remains valid for prediction.

Answer

Assumes the growth rate remains constant at 6% annually and that external conditions do not change; this assumption is critical for the model to fit the observed historical data accurately from 2020 to 2024.

Question 8

A bacterial culture doubles every 3 hours. The table records the population at selected times: Hour 0: 500 cells, Hour 3: 1,000 cells, Hour 6: 2,000 cells, Hour 9: 4,000 cells. If P(t) represents the population after t hours, which function models this data?

$P(t) = 500 \cdot 2^{t/3}$

$P(t) = 500 + 500t$

$P(t) = 500 \cdot 2^t$

$P(t) = 500 \cdot 3^{t/3}$

Accepted Answer

$$P(t) = 500 \cdot 2^{t/3}$$

Answer

$$P(t) = 500 + 500t$$

Answer

$$P(t) = 500 \cdot 2^t$$

Answer

$$P(t) = 500 \cdot 3^{t/3}$$

Question 9

A radioactive isotope's mass is recorded at the end of each year. Which statement correctly describes the relationship between consecutive terms of this sequence of masses?

Accepted Answer

Each term is 0.75 times the previous term, forming a geometric sequence with ratio 0.75.

Answer

Each term decreases by 20 grams, forming an arithmetic sequence with common difference −20.

Answer

Each term is multiplied by 1.25, indicating the mass increases by a factor of 1.25.

Answer

The ratio between consecutive terms varies, so the sequence is neither arithmetic nor geometric.

Question 10

A video goes viral with views following a geometric pattern. The table records daily views: Day 1: 500 views, Day 2: 1,500 views, Day 3: 4,500 views, Day 4: 13,500 views. If g_n denotes the views on day n, which formula correctly expresses g_n?

$g_n = 500 \cdot 3^{n-1}$

$g_n = 500 \cdot 3^n$

$g_n = 500 + 1000(n-1)$

$g_n = 1000n - 500$

Accepted Answer

$$g_n = 500 \cdot 3^{n-1}$$

Answer

$$g_n = 500 \cdot 3^n$$

Answer

$$g_n = 500 + 1000(n-1)$$

Answer

$$g_n = 1000n - 500$$

AP Precalculus Unit 2 Practice Test

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