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AP Precalculus Quiz

AP Precalculus Quiz: Rates Of Change

Practice Rates Of Change in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

The average rate of change of the function f(x)=x2f(x)=x^2f(x)=x2 is 6. If the interval is [1,b][1, b][1,b], what is the value of bbb?

Select an answer to continue

What this quiz covers

This quiz focuses on Rates Of Change, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The average rate of change of the function f(x)=x2f(x)=x^2f(x)=x2 is 6. If the interval is [1,b][1, b][1,b], what is the value of bbb?

  1. 3
  2. 4
  3. 5 (correct answer)
  4. 6

Explanation: The average rate of change is given by f(b)−f(1)b−1=6\frac{f(b) - f(1)}{b - 1} = 6b−1f(b)−f(1)​=6. Substituting the function, we get b2−12b−1=6\frac{b^2 - 1^2}{b - 1} = 6b−1b2−12​=6. Factoring the numerator gives (b−1)(b+1)b−1=6\frac{(b-1)(b+1)}{b-1} = 6b−1(b−1)(b+1)​=6. Since b≠1b \neq 1b=1, we can cancel the (b−1)(b-1)(b−1) terms, leaving b+1=6b+1 = 6b+1=6. Solving for bbb gives b=5b=5b=5.

Question 2

Let f(x)=−x3+12x+1f(x) = -x^3 + 12x + 1f(x)=−x3+12x+1. On which of the following intervals is f(x)f(x)f(x) increasing at a decreasing rate?

  1. (−∞,−2)(-\infty, -2)(−∞,−2)
  2. (−2,0)(-2, 0)(−2,0)
  3. (0,2)(0, 2)(0,2) (correct answer)
  4. (2,∞)(2, \infty)(2,∞)

Explanation: The function is increasing when its rate of change is positive. The rate of change is decreasing when the function is concave down. For a polynomial, this can be analyzed by finding where the function's graph goes up (increasing) and where it is shaped like an upside-down 'U' (concave down). The function increases between its local minimum and maximum, which occur at x=−2x=-2x=−2 and x=2x=2x=2. The point of inflection is at x=0x=0x=0. The graph is concave down for x>0x>0x>0. The interval where the function is both increasing and concave down is (0,2)(0, 2)(0,2).

Question 3

Let f(x)=2x2+1f(x) = 2x^2 + 1f(x)=2x2+1. The rate of change at x=1x=1x=1 is to be approximated by the average rate of change on the interval [1,1.1][1, 1.1][1,1.1]. What is this approximation?

  1. 4.2 (correct answer)
  2. 4
  3. 0.42
  4. 2.2

Explanation: To find the average rate of change on [1,1.1][1, 1.1][1,1.1], we calculate f(1.1)=2(1.1)2+1=2(1.21)+1=2.42+1=3.42f(1.1) = 2(1.1)^2 + 1 = 2(1.21) + 1 = 2.42 + 1 = 3.42f(1.1)=2(1.1)2+1=2(1.21)+1=2.42+1=3.42 and f(1)=2(1)2+1=3f(1) = 2(1)^2 + 1 = 3f(1)=2(1)2+1=3. The average rate of change is 3.42−31.1−1=0.420.1=4.2\frac{3.42 - 3}{1.1 - 1} = \frac{0.42}{0.1} = 4.21.1−13.42−3​=0.10.42​=4.2. This value is an approximation of the instantaneous rate of change at x=1x=1x=1.

Question 4

The population of a species of fish in a lake is modeled by the polynomial function P(t)P(t)P(t), where ttt is the time in years since the beginning of 2010. The average rate of change of P(t)P(t)P(t) over the interval [2,5][2, 5][2,5] is −150-150−150.

Based on the model, which of the following is the correct interpretation of the average rate of change?

  1. From the beginning of 2012 to the beginning of 2015, the fish population decreased on average by 150 fish per year. (correct answer)
  2. From the beginning of 2012 to the beginning of 2015, the total fish population decreased by 150 fish.
  3. At the beginning of 2015, the fish population was 150 less than it was at the beginning of 2012.
  4. From the beginning of 2012 to the beginning of 2015, the fish population increased on average by 150 fish per year.

Explanation: The average rate of change describes how one quantity changes, on average, with respect to another. The interval [2,5][2, 5][2,5] corresponds to the time from the start of 2012 to the start of 2015. A negative rate of −150-150−150 indicates an average decrease of 150 fish per year over that period.

Question 5

For which of the following intervals [a,b][a, b][a,b] is the average rate of change of the function p(x)=x2−2xp(x) = x^2 - 2xp(x)=x2−2x equal to 4?

  1. [0,4][0, 4][0,4]
  2. [1,5][1, 5][1,5] (correct answer)
  3. [2,3][2, 3][2,3]
  4. [−1,2][-1, 2][−1,2]

Explanation: The average rate of change is p(b)−p(a)b−a=4\frac{p(b)-p(a)}{b-a}=4b−ap(b)−p(a)​=4. This simplifies to (b2−2b)−(a2−2a)b−a=(b2−a2)−2(b−a)b−a=b+a−2\frac{(b^2-2b)-(a^2-2a)}{b-a} = \frac{(b^2-a^2)-2(b-a)}{b-a} = b+a-2b−a(b2−2b)−(a2−2a)​=b−a(b2−a2)−2(b−a)​=b+a−2. We need b+a−2=4b+a-2 = 4b+a−2=4, which means b+a=6b+a=6b+a=6. Of the given choices, only the interval [1,5][1, 5][1,5] has endpoints that sum to 6.

Question 6

Consider the function f(x)=1(x−3)2f(x) = \frac{1}{(x-3)^2}f(x)=(x−3)21​. How does the rate of change of f(x)f(x)f(x) behave as xxx approaches 3?

  1. The rate of change approaches zero from both the left and the right sides of x=3x = 3x=3.
  2. The rate of change increases without bound as xxx approaches 3 from the left and decreases without bound as xxx approaches 3 from the right. (correct answer)
  3. The rate of change decreases without bound as xxx approaches 3 from the left and increases without bound as xxx approaches 3 from the right.
  4. The rate of change approaches a large positive value from the left and a large negative value from the right of x=3x = 3x=3.

Explanation: The graph of f(x)f(x)f(x) has a vertical asymptote at x=3x=3x=3, with f(x)→∞f(x) \to \inftyf(x)→∞ from both sides. To the left of the asymptote (x<3x<3x<3), the function is increasing at an increasingly rapid pace, so its rate of change (slope) increases without bound. To the right of the asymptote (x>3x>3x>3), the function is decreasing at an increasingly rapid pace, so its rate of change (slope) is negative and decreases without bound.

Question 7

The polynomial function g(x)g(x)g(x) is decreasing on the interval (−∞,2)(-\infty, 2)(−∞,2) and increasing on the interval (2,∞)(2, \infty)(2,∞). Which of the following must be true?

  1. The function g(x)g(x)g(x) has a local maximum at x=2x=2x=2, and its rate of change switches from positive to negative.
  2. The function g(x)g(x)g(x) has a local maximum at x=2x=2x=2, and its rate of change switches from negative to positive.
  3. The function g(x)g(x)g(x) has a local minimum at x=2x=2x=2, and its rate of change switches from negative to positive. (correct answer)
  4. The function g(x)g(x)g(x) has a local minimum at x=2x=2x=2, and its rate of change switches from positive to negative.

Explanation: A function has a local minimum at a point where it changes from decreasing to increasing. A decreasing function has a negative rate of change, and an increasing function has a positive rate of change. Therefore, at the local minimum at x=2x=2x=2, the rate of change must switch from negative to positive.

Question 8

What is the average rate of change of the polynomial function f(x)=x3−2x+1f(x) = x^3 - 2x + 1f(x)=x3−2x+1 on the interval [−1,3][-1, 3][−1,3]?

  1. 5 (correct answer)
  2. 10
  3. 20
  4. 4.5

Explanation: The average rate of change is calculated using the formula f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a}b−af(b)−f(a)​. First, evaluate the function at the endpoints of the interval: f(3)=33−2(3)+1=27−6+1=22f(3) = 3^3 - 2(3) + 1 = 27 - 6 + 1 = 22f(3)=33−2(3)+1=27−6+1=22 and f(−1)=(−1)3−2(−1)+1=−1+2+1=2f(-1) = (-1)^3 - 2(-1) + 1 = -1 + 2 + 1 = 2f(−1)=(−1)3−2(−1)+1=−1+2+1=2. Then, apply the formula: 22−23−(−1)=204=5\frac{22 - 2}{3 - (-1)} = \frac{20}{4} = 53−(−1)22−2​=420​=5.

Question 9

Referring to the context, rate of change means output change per 1-unit input. In a physics test track, a car’s speed is modeled by the polynomial p(t)=−2t2+12tp(t)= -2t^2+12tp(t)=−2t2+12t (m/s) for 0≤t≤60\le t\le 60≤t≤6. The speed rises from t=0t=0t=0 to t=3t=3t=3 and then falls from t=3t=3t=3 to t=6t=6t=6, showing acceleration then deceleration. Over t=0t=0t=0 to t=3t=3t=3, the average rate of change is p(3)−p(0)3\frac{p(3)-p(0)}{3}3p(3)−p(0)​. Over t=3t=3t=3 to t=6t=6t=6, the average rate of change is p(6)−p(3)3\frac{p(6)-p(3)}{3}3p(6)−p(3)​. A sensor delay is modeled by the rational function r(t)=10t−3r(t)=\frac{10}{t-3}r(t)=t−310​, which has a vertical asymptote at t=3t=3t=3. As ttt approaches 3, the delay grows without bound in magnitude. Far from t=3t=3t=3, the delay approaches 0 seconds. Engineers compare these rates to decide when measurements are least reliable. Based on the passage, at which interval is the car’s average rate of change in speed negative?

  1. From t=0t=0t=0 to t=3t=3t=3
  2. From t=3t=3t=3 to t=6t=6t=6 (correct answer)
  3. From t=0t=0t=0 to t=6t=6t=6
  4. From t=2t=2t=2 to t=3t=3t=3

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, the car's speed is modeled by p(t) = -2t² + 12t, which increases from t=0 to t=3 (reaching maximum at t=3) and decreases from t=3 to t=6. Choice B is correct because the average rate of change from t=3 to t=6 is [p(6)-p(3)]/3 = [0-18]/3 = -6, which is negative, indicating the car is decelerating. Choice A is incorrect because from t=0 to t=3, the rate is [p(3)-p(0)]/3 = [18-0]/3 = 6, which is positive. Encourage students to calculate average rates systematically using the formula (f(b)-f(a))/(b-a). Watch for: confusing increasing/decreasing function values with positive/negative rates of change.

Question 10

Referring to the context, rate of change is the average change in speed per second. A car’s speed is modeled by the polynomial p(t)=−t2+10tp(t)=-t^2+10tp(t)=−t2+10t (m/s) for 0≤t≤100\le t\le 100≤t≤10. The speed increases from t=0t=0t=0 because the model starts at p(0)=0p(0)=0p(0)=0 and rises quickly. It later decreases because the negative quadratic term eventually outweighs the linear term. For example, p(3)=21p(3)=21p(3)=21 and p(7)=21p(7)=21p(7)=21, indicating the peak occurs between them. A second measurement is modeled by the rational function r(t)=50t−12+10r(t)=\frac{50}{t-12}+10r(t)=t−1250​+10, approaching 101010 as ttt increases. Near t=12t=12t=12, r(t)r(t)r(t) has a vertical asymptote and is not usable. Based on the passage, what does a negative average rate of change of p(t)p(t)p(t) on an interval indicate?

  1. The speed is decreasing over that time interval. (correct answer)
  2. The speed equals zero somewhere on that interval.
  3. The speed approaches a horizontal asymptote on that interval.
  4. The speed becomes undefined at a vertical asymptote on that interval.

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, students must interpret what a negative average rate of change means in the context of speed modeled by p(t) = -t² + 10t. Choice A is correct because a negative average rate of change means the function's output (speed) is decreasing over that time interval - the ending value is less than the starting value. Choice B is incorrect because a negative rate of change doesn't necessarily mean the function equals zero; it only indicates a decrease. Encourage students to connect mathematical concepts to physical interpretations: negative rate of change in speed means deceleration. Practice distinguishing between the sign of a function and the sign of its rate of change.

Question 11

Based on the passage, rate of change is the average change in output per unit input over an interval. In an engineering stress test, a deflection estimate is modeled by the polynomial p(x)=x3−3x2p(x)=x^3-3x^2p(x)=x3−3x2 for 0≤x≤40\le x\le 40≤x≤4, where xxx is load units. The model decreases from x=0x=0x=0 to x=2x=2x=2 and increases from x=2x=2x=2 to x=4x=4x=4, showing a turning point. A sensor correction uses the rational function r(x)=8x−2r(x)=\frac{8}{x-2}r(x)=x−28​, which has a vertical asymptote at x=2x=2x=2. Near x=2x=2x=2, the correction becomes extremely large in magnitude and destabilizes readings. Far from x=2x=2x=2, the correction approaches 0 and becomes small. Technicians compare behavior near the turning point and near the asymptote. Referring to the context, how does the polynomial’s rate-related behavior at x=2x=2x=2 differ from the rational’s behavior there?

  1. Both become unbounded at x=2x=2x=2, so neither has a usable rate
  2. The polynomial stays finite and changes direction, while the rational becomes unbounded (correct answer)
  3. The rational stays finite and changes direction, while the polynomial becomes unbounded
  4. Both approach 0 at x=2x=2x=2, so both rates must be zero there

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, the polynomial p(x) = x³ - 3x² has a turning point at x=2 (where p'(x) = 3x² - 6x = 0), while the rational function r(x) = 8/(x-2) has a vertical asymptote there. Choice B is correct because at x=2, the polynomial p(2) = 8 - 12 = -4 remains finite and simply changes from decreasing to increasing, while r(x) becomes unbounded as x approaches 2. Choice C incorrectly reverses which function becomes unbounded. Encourage students to distinguish between turning points (finite) and asymptotes (unbounded). Practice analyzing critical points versus discontinuities. Watch for: confusing polynomial extrema with rational asymptotes.

Question 12

Based on the passage, rate of change is the average change in a quantity per unit change of the input. In a physics sprint test, a cart’s speed is modeled by the polynomial p(t)=−t2+8tp(t)= -t^2+8tp(t)=−t2+8t (m/s) for 0≤t≤80\le t\le 80≤t≤8. The cart speeds up until mid-run and then slows due to friction. The average rate of change on [0,4][0,4][0,4] is p(4)−p(0)4\frac{p(4)-p(0)}{4}4p(4)−p(0)​, and on [4,8][4,8][4,8] it is p(8)−p(4)4\frac{p(8)-p(4)}{4}4p(8)−p(4)​. A timing glitch is modeled by the rational function r(t)=4t−4r(t)=\frac{4}{t-4}r(t)=t−44​ with a vertical asymptote at t=4t=4t=4. Near 4 seconds, the glitch spikes and makes readings unreliable. Away from 4 seconds, the glitch approaches 0 and becomes minor. Referring to the context, which statement correctly compares the cart’s average rates on [0,4][0,4][0,4] and [4,8][4,8][4,8]?

  1. They are both negative because the speed is always decreasing
  2. They are both positive because the speed is always increasing
  3. The first is positive, and the second is negative (correct answer)
  4. The first is zero, and the second is positive

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, the polynomial p(t) = -t² + 8t models cart speed, which has a maximum at t=4 (found by setting p'(t) = -2t + 8 = 0). Choice C is correct because on [0,4], the average rate is [p(4)-p(0)]/4 = [16-0]/4 = 4 (positive, speeding up), while on [4,8], the rate is [p(8)-p(4)]/4 = [0-16]/4 = -4 (negative, slowing down). Choice A is incorrect because the speed increases then decreases, not always decreasing. Encourage students to calculate average rates systematically and connect them to physical motion. Practice identifying turning points in quadratic functions. Watch for: assuming monotonic behavior without checking.

Question 13

Referring to the context, rate of change is the change in output divided by the change in input. In an economics simulation, daily profit (in thousands of dollars) from producing xxx units is modeled by the polynomial p(x)=−x2+10x−9p(x)= -x^2+10x-9p(x)=−x2+10x−9. Profit increases for low production and decreases after overproduction raises costs. The average rate of change from x=1x=1x=1 to x=4x=4x=4 is p(4)−p(1)3\frac{p(4)-p(1)}{3}3p(4)−p(1)​. A separate efficiency penalty is modeled by the rational function r(x)=12x−5r(x)=\frac{12}{x-5}r(x)=x−512​, which has a vertical asymptote at x=5x=5x=5. As production approaches 5 units, the penalty becomes extremely large in magnitude. For very large xxx, the penalty approaches 0, but it remains negative when x>5x>5x>5. Managers compare these behaviors to decide safe production targets. Based on the passage, what does the asymptote of r(x)r(x)r(x) indicate about the rate-related behavior near x=5x=5x=5?​

  1. The penalty becomes 0 at x=5x=5x=5, so change stops there
  2. The penalty’s magnitude grows without bound as xxx approaches 5 (correct answer)
  3. The penalty reaches its maximum finite value exactly at x=5x=5x=5
  4. The penalty equals the polynomial profit at x=5x=5x=5 by definition

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, the rational function r(x) = 12/(x-5) has a vertical asymptote at x=5, meaning the denominator equals zero there. Choice B is correct because as x approaches 5, the denominator approaches 0, causing the penalty's magnitude to grow without bound - this represents an infinite rate of change near the asymptote. Choice A is incorrect because r(5) is undefined, not zero; the function cannot equal zero at its asymptote. Encourage students to analyze vertical asymptotes as points where rational functions exhibit extreme behavior. Practice recognizing that asymptotes indicate unbounded growth, not maximum values or zeros. Watch for: misunderstanding what happens at vertical asymptotes.

Question 14

A car’s velocity is modeled by the polynomial p(t)=−t3+6t2p(t)= -t^3+6t^2p(t)=−t3+6t2 (m/s) for 0≤t≤60\le t\le 60≤t≤6. Rate of change means the average change in velocity per second over an interval. From t=0t=0t=0 to t=2t=2t=2, the car speeds up as p(t)p(t)p(t) rises. From t=2t=2t=2 to t=4t=4t=4, the increase slows because the added velocity each second shrinks. From t=4t=4t=4 to t=6t=6t=6, velocity decreases, indicating deceleration. Engineers also track sensor error with r(t)=12t−3+8r(t)=\frac{12}{t-3}+8r(t)=t−312​+8, which has a vertical asymptote at t=3t=3t=3. Near t=3t=3t=3, small time changes cause large jumps in r(t)r(t)r(t). Far from t=3t=3t=3, r(t)r(t)r(t) levels toward 8, so its rate of change diminishes. Based on the passage, what is the average rate of change of p(t)p(t)p(t) on [4,6][4,6][4,6]?​

  1. −8-8−8 m/s per s
  2. −12-12−12 m/s per s
  3. −16-16−16 m/s per s (correct answer)
  4. 161616 m/s per s

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. To find the average rate of change on [4,6], we calculate [p(6) - p(4)]/(6-4) where p(t) = -t³+6t². At t=4: p(4) = -64+96 = 32; at t=6: p(6) = -216+216 = 0, giving average rate = (0-32)/2 = -16 m/s per s. Choice C is correct because the calculation yields a decrease of 32 m/s over 2 seconds, which equals -16 m/s per second. Choice D is incorrect because it gives the magnitude without the negative sign, missing that this represents deceleration. Encourage students to maintain negative signs when computing rates of decreasing functions. Practice calculating rates of change for intervals where functions decrease.

Question 15

Based on the passage, rate of change is the average change in speed per second over a chosen interval. A car’s speed is modeled by the polynomial p(t)=−t2+6t+8p(t)=-t^2+6t+8p(t)=−t2+6t+8 (m/s) for 0≤t≤60\le t\le 60≤t≤6. The model increases at first, since p(0)=8p(0)=8p(0)=8 and p(2)=16p(2)=16p(2)=16 show acceleration. It then decreases after the peak because the negative quadratic term reduces the speed. For example, p(4)=16p(4)=16p(4)=16 and p(6)=8p(6)=8p(6)=8 show deceleration later. A separate drag-limited model is r(t)=12t−7+6r(t)=\frac{12}{t-7}+6r(t)=t−712​+6, approaching 666 as ttt increases. The rational model is undefined at t=7t=7t=7 due to a vertical asymptote. Referring to the context, how does the rate of change of r(t)r(t)r(t) differ near its vertical asymptote compared with large ttt?​

  1. It changes very rapidly near t=7t=7t=7 but changes slowly as ttt becomes large. (correct answer)
  2. It changes slowly near t=7t=7t=7 but changes rapidly as ttt becomes large.
  3. It is constant near t=7t=7t=7 and constant for large ttt.
  4. It reaches a maximum at t=7t=7t=7 because asymptotes are turning points.

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, the rational function r(t) = 12/(t-7) + 6 has a vertical asymptote at t=7, causing dramatic changes in function values near this point. Choice A is correct because near the vertical asymptote at t=7, small changes in t produce large changes in r(t), resulting in rapid rates of change, while as t becomes large, r(t) approaches 6 and changes slowly. Choice B incorrectly reverses this behavior - rational functions change most rapidly near their vertical asymptotes, not at large t values. Encourage students to visualize the graph of rational functions near vertical asymptotes versus their horizontal asymptotic behavior. Practice analyzing how proximity to asymptotes affects rates of change.

Question 16

Referring to the context, rate of change means average change in speed per second over a stated interval. A car’s speed is modeled by the polynomial p(t)=−0.5t2+5t+4p(t)=-0.5t^2+5t+4p(t)=−0.5t2+5t+4 (m/s) for 0≤t≤100\le t\le 100≤t≤10. The speed increases early, since values like p(0)=4p(0)=4p(0)=4 and p(4)=16p(4)=16p(4)=16 show acceleration. Later, the speed decreases because the negative quadratic term reduces speed growth. A second sensor follows the rational model r(t)=18t−6+9r(t)=\frac{18}{t-6}+9r(t)=t−618​+9, which approaches 999 as ttt increases. The rational model has a vertical asymptote at t=6t=6t=6, where the expression is undefined. Based on the passage, how does the rate of change of r(t)r(t)r(t) behave as ttt gets large?

  1. It approaches 0 because the function levels off toward its horizontal asymptote. (correct answer)
  2. It increases without bound because the numerator stays constant.
  3. It becomes infinite because the function approaches the vertical asymptote.
  4. It stays constant because rational functions have constant average rates of change.

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, the rational function r(t) = 18/(t-6) + 9 approaches the horizontal asymptote y = 9 as t increases, which affects its rate of change behavior. Choice A is correct because as t gets large, r(t) approaches its horizontal asymptote of 9, meaning the function levels off and its rate of change approaches 0. Choice C is incorrect because it confuses the behavior near the vertical asymptote (at t=6) with the behavior as t approaches infinity. Encourage students to visualize how rational functions flatten out as they approach horizontal asymptotes. Practice analyzing long-term behavior of rational functions and connecting this to diminishing rates of change.

Question 17

Referring to the context, rate of change is measured by the slope between two points on a model. In a biology tank, nutrient level is approximated by the polynomial p(t)=−0.5t2+3t+2p(t)= -0.5t^2+3t+2p(t)=−0.5t2+3t+2 for 0≤t≤60\le t\le 60≤t≤6 (days). It increases early and then decreases after the peak as nutrients get consumed. The average rate of change from t=2t=2t=2 to t=4t=4t=4 is p(4)−p(2)2\frac{p(4)-p(2)}{2}2p(4)−p(2)​. A probe error is modeled by the rational function r(t)=3t−3r(t)=\frac{3}{t-3}r(t)=t−33​ with a vertical asymptote at t=3t=3t=3. Very close to day 3, the error becomes extremely large and can mask real trends. Far from day 3, the error approaches 0 and becomes negligible. Based on the passage, which interpretation of the asymptote is mathematically accurate?

  1. At t=3t=3t=3, r(t)r(t)r(t) has a highest value that the error cannot exceed
  2. As ttt nears 3, r(t)r(t)r(t) grows without bound in magnitude (correct answer)
  3. At t=3t=3t=3, r(t)r(t)r(t) equals 0, so the error disappears instantly
  4. As ttt nears 3, r(t)r(t)r(t) becomes constant, so the error stabilizes

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, the rational function r(t) = 3/(t-3) has a vertical asymptote at t=3, meaning the function becomes unbounded there. Choice B is correct because as t approaches 3, the denominator approaches 0, causing |r(t)| to grow without bound - this is the mathematical definition of a vertical asymptote. Choice C is incorrect because r(3) is undefined (division by zero), not equal to 0; the error becomes infinite, not zero. Encourage students to understand asymptotes as points of unbounded behavior, not zeros or maxima. Practice recognizing the mathematical meaning of vertical asymptotes. Watch for: confusing undefined values with zero values.

Question 18

Let f(x)=x2−6x+5f(x) = x^2 - 6x + 5f(x)=x2−6x+5. How does the average rate of change of fff over the interval [4,6][4, 6][4,6] compare to the average rate of change of fff over the interval [1,3][1, 3][1,3]?

  1. The average rate of change over [4,6][4, 6][4,6] is greater than the average rate of change over [1,3][1, 3][1,3]. (correct answer)
  2. The average rate of change over [4,6][4, 6][4,6] is less than the average rate of change over [1,3][1, 3][1,3].
  3. The average rates of change over the two intervals are equal.
  4. The relationship cannot be determined from the information given.

Explanation: First, calculate the average rate of change (AROC) for [4,6][4, 6][4,6]: f(6)=5f(6) = 5f(6)=5, f(4)=−3f(4) = -3f(4)=−3. AROC = 5−(−3)6−4=82=4\frac{5 - (-3)}{6 - 4} = \frac{8}{2} = 46−45−(−3)​=28​=4. Next, calculate the AROC for [1,3][1, 3][1,3]: f(3)=−4f(3) = -4f(3)=−4, f(1)=0f(1) = 0f(1)=0. AROC = −4−03−1=−42=−2\frac{-4 - 0}{3 - 1} = \frac{-4}{2} = -23−1−4−0​=2−4​=−2. Since 4>−24 > -24>−2, the average rate of change over [4,6][4, 6][4,6] is greater.

Question 19

The average rates of change of a polynomial function q(x)q(x)q(x) over consecutive equal-length input-value intervals can be described by a linear function with a non-zero slope. Which of the following must be the most specific classification for q(x)q(x)q(x)?

  1. q(x)q(x)q(x) is a linear function.
  2. q(x)q(x)q(x) is a quadratic function. (correct answer)
  3. q(x)q(x)q(x) is a cubic function.
  4. The degree of q(x)q(x)q(x) cannot be determined.

Explanation: For a polynomial, if the average rates of change over consecutive equal-length intervals form a linear pattern (i.e., the second differences of the function values are constant and non-zero), the function must be quadratic. A linear function would have constant average rates of change, and a cubic function's average rates of change would form a quadratic pattern.

Question 20

The polynomial function p(x)p(x)p(x) has a single local maximum at x=cx=cx=c. Which statement best describes the function's rate of change immediately around this point?

  1. The rate of change is positive for x<cx < cx<c and negative for x>cx > cx>c. (correct answer)
  2. The rate of change is negative for x<cx < cx<c and positive for x>cx > cx>c.
  3. The rate of change is positive for both x<cx < cx<c and x>cx > cx>c.
  4. The rate of change is negative for both x<cx < cx<c and x>cx > cx>c.

Explanation: A local maximum occurs where a function stops increasing and starts decreasing. An increasing function has a positive rate of change, and a decreasing function has a negative rate of change. Therefore, immediately to the left of the maximum (x<cx < cx<c), the rate of change is positive, and immediately to the right (x>cx > cx>c), the rate of change is negative.