Drill 1 · Math · Rates of Change
AP Precalculus: Rates of Change (Drill 1) is a Math practice drill covering Rates of Change. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
Rates of change questions on the AP® Precalculus exam test how to compute and interpret average rates of change from tables and equations, identify function types from patterns in data, and distinguish between increasing at an increasing rate versus increasing at a decreasing rate. This drill covers average rate of change with units and context, identifying linear, quadratic, and exponential models from tables, and interpreting concavity language in real-world settings.
Question 1. The table below shows the height h (in feet) of a balloon at selected times t (in minutes).
| t (min) | 0 | 2 | 4 | 6 |
|---|---|---|---|---|
| h (ft) | 50 | 74 | 98 | 122 |
Explanation: Choice C is correct. Average rate of change = (h(6) − h(2)) / (6 − 2) = (122 − 74) / 4 = 48 / 4 = 12 feet per minute. This means the balloon rises an average of 12 feet for each minute elapsed between t = 2 and t = 6. Choice A is incorrect because 24 feet per minute results from dividing the total change (48) by 2, the length of one sub-interval, rather than by 4, the full interval length. Choice B is incorrect because 48 feet per minute confuses total change in height with average rate of change; the total change must be divided by the interval length to obtain a rate. Choice D is incorrect because 12 feet is not a rate; it omits "per minute" and misidentifies the result as a height rather than a rate of change.
Question 2. The table below gives values of a function f at equally spaced values of x.
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| f(x) | 3 | 6 | 12 | 24 | 48 |
Explanation: Choice D is correct. The ratios of consecutive outputs are 6/3 = 2, 12/6 = 2, 24/12 = 2, 48/24 = 2, a constant ratio at equally spaced x-values is the defining feature of an exponential function. Choice A is incorrect because the first differences (3, 6, 12, 24) are not constant; they double each step. Choice B is incorrect because constant second differences identify a quadratic model; the second differences here are 3, 6, 12, not constant. Choice C is incorrect because increasing first differences are consistent with both quadratic and exponential models; only the constant ratio distinguishes this data as exponential.
Question 3. The table below gives values of a function g at equally spaced values of x.
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| g(x) | 4 | 7 | 12 | 19 | 28 |
Explanation: Choice C is correct. First differences: 7−4 = 3, 12−7 = 5, 19−12 = 7, 28−19 = 9. These are not constant, ruling out a linear model. Second differences: 5−3 = 2, 7−5 = 2, 9−7 = 2. The second differences are constant (all equal 2), which is the defining feature of a quadratic model. Choice A is incorrect because the first differences (3, 5, 7, 9) are not constant, the output increases by a different amount each step. Choice B is incorrect because always increasing describes linear, quadratic, and exponential functions alike; it does not identify the model type. Choice D is incorrect because increasing first differences are consistent with both quadratic and exponential models; only constant second differences confirm a quadratic model specifically.
Question 4. A cyclist begins a ride and her speed increases throughout the first 10 minutes. The table below shows her speed s (in miles per hour) at selected times t (in minutes).
| t (min) | 0 | 2 | 4 | 6 | 8 | 10 |
|---|---|---|---|---|---|---|
| s (mph) | 0 | 6 | 10 | 13 | 15 | 16 |
Explanation: Choice A is correct. The average rate of change of s over each 2-minute interval: (6−0)/2 = 3, (10−6)/2 = 2, (13−10)/2 = 1.5, (15−13)/2 = 1, (16−15)/2 = 0.5. These values decrease (3, 2, 1.5, 1, 0.5), meaning each additional 2 minutes produces a smaller gain in speed, the speed is increasing at a decreasing rate (concave down). Choice B is incorrect because speed values getting larger only confirms speed is increasing, not the rate at which it is increasing; a student must examine successive rates of change, not output values alone. Choice C is incorrect because a constant rate of increase would require equal speed gains in equal time intervals, but the gains (6, 4, 3, 2, 1) are shrinking. Choice D is incorrect because the maximum speed reached has no bearing on whether the rate of increase is increasing or decreasing.
Question 5. The function f is given by f(x) = 2x2 + 3x. What is the average rate of change of f on the interval [0, 3]?
Explanation: Choice C is correct. Compute f(0) and f(3): f(0) = 2(0)2 + 3(0) = 0. f(3) = 2(3)2 + 3(3) = 2(9) + 9 = 18 + 9 = 27. Average rate of change = (f(3) − f(0)) / (3 − 0) = (27 − 0) / 3 = 9. Choice A is incorrect because 27 is the value of f(3) itself, the student correctly computed the output but forgot to subtract f(0) and divide by the interval length to find the average rate of change. Choice B is incorrect because 3 results from treating the denominator as 32 = 9 rather than the interval length 3 − 0 = 3, giving 27/9 = 3, a sign that the student confused squaring the endpoint with computing interval length. Choice D is incorrect because 5 results from computing f(3) as 2(3) + 3(3) = 6 + 9 = 15, using 2(3) instead of 2(32) = 18, and then dividing by 3: 15/3 = 5, which is the classic error of forgetting to apply the exponent before multiplying.