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Q 01 / 20
Easy
Which of the following best describes an exponential function ?
A A function where the variable is in the base: $f(x) = x^2$
B A function where the variable is in the exponent: $f(x) = 2^x$
C A function that forms a straight line
D A function where each term increases by a fixed amount
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Q 02 / 20 Easy
A population of bacteria doubles every hour . If the initial count is 500, which function models the population after $t$ hours?
A $P(t) = 500 + 2t$
B $P(t) = 500 \cdot t^2$
C $P(t) = 500 \cdot 2^t$
D $P(t) = 2 \cdot 500^t$
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Q 03 / 20 Easy
If a savings account grows at 5% per year , what is the base of the exponential function?
A 5
B 0.05
C 0.95
D 1.05
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Q 04 / 20 Easy
Evaluate: $f(x) = 3 \cdot 2^x$ at $x = 4$.
A 24
B 36
C 48
D 96
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Q 05 / 20 Easy
A car's value decreases by 15% each year . If the original price is $20,000, which function models the car's value $V$ after $t$ years?
A $V(t) = 20000 \cdot 1.15^t$
B $V(t) = 20000 \cdot 0.85^t$
C $V(t) = 20000 - 15t$
D $V(t) = 20000 \cdot 0.15^t$
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Q 06 / 20 Medium
The function below models a city's population in thousands, $t$ years after 2003:
$$P(t) = 260(1.04)^{\left(\frac{6}{4}\right)t}$$
According to this model, the population increases by 4% every $n$ months . What is $n$?
ANCHOR-03 SCALE: The exponent $\frac{6}{4} \cdot t$ means "how many periods of $\frac{4}{6}$ years" have passed. Convert years β months.
A 8
B 12
C 18
D 72
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Q 07 / 20 Medium
The function $P(t) = 180(1.06)^{2t}$ models a population (in thousands). How often does it grow by 6% ?
A Every year
B Every 2 years
C Every 6 months (half a year)
D Every 12 months
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Q 08 / 20 Medium
Which of the following is equivalent to $2^{3t}$?
A $6^t$
B $8^t$
C $2^3 \cdot t$
D $3 \cdot 2^t$
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Q 09 / 20 Medium
A town's population is modeled by $P(t) = 5000 \cdot (1.03)^{t/5}$. What does the "5" in the exponent tell you?
A The population grows 5% per year
B The initial population is 5,000
C The population grows 3% every 5 years
D The growth rate is 5% every 3 years
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Q 10 / 20 Medium
A radioactive substance has a half-life of 8 years . Starting with 200 grams, which function models the remaining mass $M$ after $t$ years?
A $M(t) = 200 \cdot 2^{t/8}$
B $M(t) = 200 \cdot 0.5^{8t}$
C $M(t) = 200 - 8t$
D $M(t) = 200 \cdot (0.5)^{t/8}$
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Q 11 / 20 Medium
The function $f(x) = 400 \cdot (0.92)^x$ models the population of a species after $x$ years. What is the annual percent decrease ?
A 92%
B 8%
C 0.92%
D 0.08%
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Q 12 / 20 Medium
A function is given as $f(t) = 120 \cdot (1.05)^{t/12}$. What is the monthly growth rate?
A 5% every 12 months (i.e., every year)
B 5% every month
C 12% every 5 months
D 60% per year
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Q 13 / 20 Hard
The function $P(t) = 500(1.02)^{(4/3)t}$ models a population where $t$ is in years. The population grows by 2% every $n$ months. Find $n$.
ANCHOR-03 SCALE: The period is $\frac{3}{4}$ of a year. Convert: $\frac{3}{4} \times 12 = ?$ months.
A 4 months
B 9 months
C 12 months
D 16 months
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Q 14 / 20 Hard
Which expression is equivalent to $(1.04)^{(6/4)t}$? (Think: rewrite in the form $b^t$)
A $(1.04)^{2.5t}$
B $(1.04)^{0.67t}$
C $\left((1.04)^{1.5}\right)^t$
D $1.04 \cdot 1.5^t$
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Q 15 / 20 Hard
Two functions are given: $f(t) = 100 \cdot (1.1)^t \quad$ and $\quad g(t) = 100 \cdot (1.1)^{2t}$ Compared to $f$, the function $g$ reaches $200$ (double) in:
A Half the time
B The same time
C Twice the time
D 10% less time
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Q 16 / 20 Hard
The population model is $P(t) = 1000 \cdot b^t$ where $P(3) = 1728$. What is the value of $b$?
A 1.2
B 1.2 (since $1.2^3 = 1.728$, scaled by 1000)
C 1.728
D 3
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Q 17 / 20 Hard
A quantity grows by 20% every 3 months. Which function correctly models this, where $t$ is in years ?
A $f(t) = (1.2)^{t/3}$
B $f(t) = (1.2)^{3t}$
C $f(t) = (1.2)^{t/12}$
D $f(t) = (1.2)^{4t}$
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Q 18 / 20 Hard
If $f(x) = a \cdot b^x$ passes through $(0, 4)$ and $(2, 36)$, what are the values of $a$ and $b$?
A $a = 2,\; b = 9$
B $a = 4,\; b = 9$
C $a = 4,\; b = 3$
D $a = 36,\; b = 4$
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Q 19 / 20 Hard
The function $P(t) = 260(1.04)^{(6/4)t}$ is rewritten as $P(t) = 260 \cdot k^t$. What is the approximate value of $k$?
ANCHOR-05 TRAP: Compute $(1.04)^{6/4} = (1.04)^{1.5}$ β this is the annual multiplier.
A 1.04
B β 1.0612
C β 1.08
D 1.5
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Q 20 / 20 Hard
A model is $P(t) = 400(1.03)^{(12/3)t}$ where $t$ is in years. By what percent does the population increase every 3 months ? And what is the equivalent annual growth rate?
$$P(t) = 400(1.03)^{4t}$$
A 3% quarterly; 12% annually
B 3% quarterly; 3% annually
C 3% quarterly; β 12.55% annually
D 12% quarterly; 3% annually
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