Task Sets: Interpret exponential function properties - Common Core Math - HS Functions

A

Recall / Warm-Up

1.

Simplify: $(1.05)^{12t}$ using the exponent rule $(a^m)^n = a^{mn}$ .

A.

$((1.05)^{12})^t$

B.

$(1.05)^{12} \cdot t$

C.

$(1.05)^{t} + 12$

D.

$12 \cdot (1.05)^t$

2.

For $f(t) = 3000 \cdot (1.07)^t$ , what is the initial value (the value when $t = 0$ )?

A.

$1.07$

B.

$1$

C.

$3000$

D.

$0.07$

3.

Which base value indicates exponential decay?

A.

$b = 1.5$

B.

$b = 1$

C.

$b = 0.75$

D.

$b = -0.5$

B

Fluency Practice

C

Varied Practice

D

Word Problems

1.

An investment account starts with $5,000 and grows at 0.4% interest per month. The balance is modeled by $B(t) = 5000 \cdot (1.004)^{12t}$ , where $t$ is in years.

1.

Rewrite $B(t)$ in the form $B(t) = 5000 \cdot b^t$ and select the best description of the annual rate.

A.

Annual growth factor $\approx 1.049$ ; about 4.9% annual growth (not $12 \times 0.4\% = 4.8\%$ due to compounding).

B.

Annual growth factor $= 1.048$ ; exactly 4.8% annual growth ( $12 \times 0.4\%$ ).

C.

Annual growth factor $= 0.004$ ; 0.4% annual growth.

D.

The function cannot be rewritten in the form $b^t$ .

2.

Using the original monthly model $B(t) = 5000 \cdot (1.004)^{12t}$ , what is the balance after 2 years (in dollars)? Round to the nearest dollar.

2.

Two towns have populations modeled by $A(t) = 2000 \cdot (1.10)^t$ and $B(t) = 50000 \cdot (1.02)^t$ , where $t$ is years from now.

Town A starts with 2,000 people and grows at 10% per year. Town B starts with 50,000 and grows at 2%. At $t = 0$ , Town B's population is how many times as large as Town A's population?

E

Error Analysis

1.

A student analyzed $f(t) = 750 \cdot (1.08)^t$ and wrote:

"The growth rate is 1.08, which means 108% growth per year. Starting at 750, the quantity more than doubles each year."

What error did the student make?

A.

The student confused the base with the initial value — $750$ is the rate, $1.08$ is the initial value.

B.

The student confused the growth factor with the growth rate. The factor is $1.08$ ; the rate is $1.08 - 1 = 0.08 = 8\%$ per year.

C.

The student correctly identified the rate. 108% growth per year is correct when the base is 1.08.

D.

The function is actually a decay function because $1.08 < 1.10$ .

2.

A student saw $g(t) = P \cdot (1.01)^{12t}$ and wrote:

"Each year, the balance grows by $12 \times 1\% = 12\%$ . So the annual growth rate is 12%."

What is wrong with this reasoning, and what is the correct annual growth rate?

A.

The reasoning is correct — 1% monthly for 12 months is 12% annually.

B.

Multiplication applies to linear growth, not exponential. The correct annual rate is $(1.01)^{12} - 1 \approx 12.68\%$ due to compounding.

C.

The correct annual rate is $1\% / 12 \approx 0.083\%$ — you divide, not multiply.

D.

The annual rate is exactly 12% — the compounding effect is negligible.

F

Challenge / Extension

1.

A radioactive element has a half-life of 8 years. Starting with 200 grams, the model is $A(t) = 200 \cdot (0.5)^{t/8}$ . What percent of the original amount remains after 20 years? Round to the nearest whole percent.

2.