AP Precalculus Exponential Functions and Modeling — Questions and Answers
Question 1: An arithmetic sequence has a_3 = 11 and a_7 = 27. What is a_20?
- 76
- 75
- 79 (Correct answer)
- 83
Correct 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.
In an arithmetic sequence, each step adds the same common difference d. The gap between a_3 and a_7 is four steps, so 4d = 27 − 11 = 16 and d = 4. Working backward two steps from a_3 gives a_1 = 11 − 8 = 3. The explicit form is a_n = 3 + 4(n − 1), so a_20 = 3 + 76 = 79.
Question 2: A geometric sequence begins 6, 18, 54, ... What is the sixth term?
- 1458 (Correct answer)
- 4374
- 486
- 1296
Correct answer: 1458
The common ratio is 3, so g_n = 6·3^(n−1). Then g_6 = 6·3^5 = 6·243 = 1458.
A geometric sequence multiplies by a constant ratio r each step. Here 18/6 = 3 and 54/18 = 3, so r = 3. The explicit rule is g_n = g_1·r^(n−1), which means the sixth term uses the exponent 5. This gives 6·243 = 1458. Using exponent 6 would give 4374, which is the seventh term.
Question 3: A town has 2400 residents, and its population decreases by 6% each year. Which function models the population P after t years?
- P(t) = 2400(1.06)^t
- P(t) = 2400 − 0.06t
- P(t) = 2400(0.06)^t
- P(t) = 2400(0.94)^t (Correct answer)
Correct 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.
Constant percent change over equal time intervals is the signature of exponential behavior. For percent decay at rate r, the base is 1 − r, which is 0.94 here. A base of 1.06 would model growth. A base of 0.06 would keep only 6% of the population each year. The linear option subtracts a fixed amount rather than a fixed percent.
Question 4: Which function is equivalent to f(t) = 5·2^(3t) written in the form a·b^t?
- 5·6^t
- 5·8^t (Correct answer)
- 15·2^t
- 125·2^t
Correct answer: 5·8^t
By the power rule for exponents, 2^(3t) = (2^3)^t = 8^t. So f(t) = 5·8^t.
The property (b^m)^t = b^(mt) lets you move a constant factor in the exponent into the base. Here 2^(3t) = (2^3)^t = 8^t. The coefficient 5 is not part of the exponent and stays the same. Multiplying the base by 3 or the coefficient by 3 does not preserve the function's values.
Question 5: 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?
Correct answer: 12
Set 200·2^(t/3) = 3200, so 2^(t/3) = 16 = 2^4. Then t/3 = 4 and t = 12 hours.
Dividing both sides by 200 gives 2^(t/3) = 16. Since 16 = 2^4, the exponents must be equal, so t/3 = 4. Therefore t = 12 hours, which is four doubling periods of 3 hours each. Check: 200 doubled four times is 400, 800, 1600, 3200.
Question 6: 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?
- y = 4·3^x (Correct answer)
- y = 4 + 8x
- y = 4x² + 8x
- y = 3·4^x
Correct 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.
Equal input intervals with a constant ratio between outputs indicate exponential behavior. The ratios 12/4, 36/12 and 108/36 are all 3. The initial value is f(0) = 4, so f(x) = 4·3^x. The linear model fails because the differences (8, 24, 72) are not constant. The quadratic model gives f(2) = 32, not 36.
Question 7: 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?
- The residuals form a clear U-shaped curve.
- The residuals increase steadily as the input increases.
- All residuals are positive.
- The residuals are scattered randomly above and below zero with no pattern. (Correct answer)
Correct 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.
If a model is appropriate, what remains after fitting should look like noise. A curved or trending residual pattern means the model is missing structure, so a different model type may fit better. All-positive residuals mean the model consistently underestimates. Random scatter around zero, with no systematic pattern, supports the model. This is how competing models are compared.
Question 8: An exponential function f(x) = a·b^x with b > 0 satisfies f(2) = 45 and f(4) = 405. What is f(5)?
- 1125
- 1215 (Correct answer)
- 3645
- 675
Correct 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.
Over an input change of 2, the output is multiplied by b². Because 405/45 = 9, b² = 9, and b = 3 since b is positive. Substituting into f(2) = a·9 = 45 gives a = 5. So f(x) = 5·3^x, and f(5) = 5·243 = 1215. Using x = 6 would give 3645.
Question 9: Let g(x) = −3·2^x + 5. Which statement about g is true?
- g has a horizontal asymptote at y = −3 and increases without bound as x → ∞.
- g has a horizontal asymptote at y = 5 and increases without bound as x → ∞.
- g has a horizontal asymptote at y = 5 and decreases without bound as x → ∞. (Correct answer)
- g has a horizontal asymptote at y = 2 and decreases without bound as x → ∞.
Correct 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 → ∞.
The parent function 2^x has a horizontal asymptote at y = 0 as x → −∞. Multiplying by −3 stretches and reflects it over the x-axis, so it becomes decreasing. Adding 5 shifts the asymptote up to y = 5. As x → −∞, −3·2^x approaches 0, so g(x) approaches 5. As x → ∞, 2^x grows without bound, so g(x) → −∞.
Question 10: 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)?
- 15
- 25
- 30
- 20 (Correct answer)
Correct 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.
Over any input interval of length 2, an exponential function multiplies its output by b². Here 40/10 = 160/40 = 4, so b² = 4, and b = 2 because b is positive. Using f(1) = a·2 = 10 gives a = 5. So f(x) = 5·2^x, and f(2) = 5·4 = 20. The value 25 would result from averaging 10 and 40, which wrongly treats the function as linear.
