Exponential Functions and Modeling Flashcards

An arithmetic sequence has a_3 = 11 and a_7 = 27. What is a_20?

Answer: 79

From a_7 − a_3 = 4d = 16, the common difference is d = 4. Then a_1 = 11 − 2(4) = 3, so a_20 = 3 + 19(4) = 79.

A geometric sequence begins 6, 18, 54, ... What is the sixth term?

Answer: 1458

The common ratio is 3, so g_n = 6·3^(n−1). Then g_6 = 6·3^5 = 6·243 = 1458.

A town has 2400 residents, and its population decreases by 6% each year. Which function models the population P after t years?

Answer: P(t) = 2400(0.94)^t

A 6% decrease each year multiplies the population by 1 − 0.06 = 0.94 every year. That gives P(t) = 2400(0.94)^t.

Which function is equivalent to f(t) = 5·2^(3t) written in the form a·b^t?

Answer: 5·8^t

By the power rule for exponents, 2^(3t) = (2^3)^t = 8^t. So f(t) = 5·8^t.

A bacteria culture starts at 200 cells and doubles every 3 hours, so N(t) = 200·2^(t/3). After how many hours will the culture reach 3200 cells?

Answer: 12

Set 200·2^(t/3) = 3200, so 2^(t/3) = 16 = 2^4. Then t/3 = 4 and t = 12 hours.

The table gives values of a function: f(0) = 4, f(1) = 12, f(2) = 36, f(3) = 108. Which model best fits the data?

Answer: y = 4·3^x

Each output is 3 times the previous one over equal input steps, so the data are exponential with initial value 4 and base 3. That gives y = 4·3^x.

A student fits an exponential model to a data set and plots the residuals. Which observation about the residual plot best supports the model being appropriate?

Answer: The residuals are scattered randomly above and below zero with no pattern.

A residual is the actual value minus the predicted value. Random scatter around zero with no pattern shows the model captures the trend in the data.

An exponential function f(x) = a·b^x with b > 0 satisfies f(2) = 45 and f(4) = 405. What is f(5)?

Answer: 1215

Since f(4)/f(2) = b² = 9, b = 3. Then a = 45/9 = 5, so f(5) = 5·3^5 = 5·243 = 1215.

Let g(x) = −3·2^x + 5. Which statement about g is true?

Answer: g has a horizontal asymptote at y = 5 and decreases without bound as x → ∞.

The vertical shift of +5 places the horizontal asymptote at y = 5, as x → −∞. The negative coefficient reflects the graph, so g(x) → −∞ as x → ∞.

For an exponential function f(x) = a·b^x with b > 0, f(1) = 10, f(3) = 40 and f(5) = 160. What is f(2)?

Answer: 20

Outputs are multiplied by 4 over each 2-unit step, so b² = 4 and b = 2. Then a = 10/2 = 5, and f(2) = 5·2² = 20.