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AP Precalculus -2.2 Linear and Exponential Change- FRQ Exam Style Questions - Effective Fall 2023

AP Precalculus -2.2 Linear and Exponential Change- FRQ Exam Style Questions – Effective Fall 2023

AP Precalculus -2.2 Linear and Exponential Change- FRQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – FRQ Exam Style Questions- All Topics

Question

 
The table gives the number of ice cream cones sold by a food vendor. On the initial day \(t = 0\) when the vendor added ice cream cones to the menu, the vendor sold 14 ice cream cones. Thirty-five days later \(t = 35\), the vendor sold 57 ice cream cones. Ten days after that \(t = 45\), the vendor sold 46 ice cream cones.
The number of ice cream cones sold can be modeled by the quadratic function \(I\) given by \(I(t) = at^2 + bt + c\), where \(I(t)\) is the number of ice cream cones sold on day \(t\).
(A)
(i) Use the given data to write three equations that can be used to find the values for constants \(a\), \(b\), and \(c\) in the expression for \(I(t)\).
(ii) Find the values for \(a\), \(b\), and \(c\) as decimal approximations.
(B)
(i) Use the given data to find the average rate of change of the number of ice cream cones sold, in cones per day, from \(t = 35\) to \(t = 45\) days. Express your answer as a decimal approximation. Show the computations that lead to your answer.
(ii) Use the average rate of change found in (i) to estimate the number of ice cream cones sold on day \(t = 40\). Show the work that leads to your answer.
(iii) Compare the estimate found in (ii) to the value given by the model, \(I(40)\). Using characteristics of the average rate of change and characteristics of the quadratic model, explain why the two estimates differ.
(C)
Explain how the range values of the function \(I\) should be limited by the context of the problem.

Most-appropriate topic codes (AP Precalculus CED):

• 1.14: Function Model Construction and Application — parts (A), (B)(iii), (C)
• 2.2: Change in Linear and Exponential Functions — parts (B)(i), (B)(ii)
• 2.6: Competing Function Model Validation — part (B)(iii)

▶️ Answer/Explanation

(A)

(i) Substituting the given data into the model \(I(t) = at^2 + bt + c\):

  • When \(t = 0\), \(I(0) = 14\): \(a(0)^2 + b(0) + c = 14 \Rightarrow c = 14\).
  • When \(t = 35\), \(I(35) = 57\): \(a(35)^2 + b(35) + c = 57 \Rightarrow 1225a + 35b + c = 57\).
  • When \(t = 45\), \(I(45) = 46\): \(a(45)^2 + b(45) + c = 46 \Rightarrow 2025a + 45b + c = 46\).

The three equations are:
\(c = 14\)
\(1225a + 35b + 14 = 57\)
\(2025a + 45b + 14 = 46\)
✅ Answer: The three equations are \(c = 14\), \(1225a + 35b = 43\), and \(2025a + 45b = 32\).

(ii) Solve the system:

  1. \(1225a + 35b = 43\)
  2. \(2025a + 45b = 32\)

Multiply equation (1) by 9 and equation (2) by 7 to eliminate \(b\): \[ 11025a + 315b = 387 \] \[ 14175a + 315b = 224 \] Subtract the first from the second: \[ 3150a = -163 \Rightarrow a = -\frac{163}{3150} \approx -0.051746\ldots \] Substitute \(a\) into \(1225a + 35b = 43\): \[ 1225(-0.051746) + 35b = 43 \Rightarrow -63.38885 + 35b = 43 \Rightarrow 35b = 106.38885 \Rightarrow b \approx 3.039681 \] Thus, \[ a \approx -0.052, \quad b \approx 3.040, \quad c = 14 \] ✅ Answer: \(\boxed{a \approx -0.052,\ b \approx 3.040,\ c = 14}\) (or \(a \approx -0.0517,\ b \approx 3.0397,\ c = 14\)).

(B)

(i) Average rate of change from \(t = 35\) to \(t = 45\): \[ \frac{I(45) – I(35)}{45 – 35} = \frac{46 – 57}{10} = \frac{-11}{10} = -1.1 \] ✅ Answer: \(\boxed{-1.1}\) cones per day.

(ii) Using the average rate of change: \[ I(40) \approx I(35) + (-1.1)(40 – 35) = 57 + (-1.1)(5) = 57 – 5.5 = 51.5 \] ✅ Answer: \(\boxed{51.5}\) cones (approximately 51 or 52 cones).

(iii) Using the model: \[ I(40) \approx -0.052(40)^2 + 3.040(40) + 14 = -0.052(1600) + 121.6 + 14 = -83.2 + 135.6 = 52.4 \] More precisely with stored values: \(I(40) \approx 52.794\).
The estimate from the average rate of change (51.5) is less than \(I(40) \approx 52.8\).
Explanation: The average rate of change gives the slope of the secant line between \(t = 35\) and \(t = 45\). Since the quadratic model has \(a < 0\), its graph is concave down. On the interval \((35, 45)\), the secant line lies below the graph. Therefore, the linear estimate using the average rate of change underestimates the actual value given by the concave-down quadratic at \(t = 40\).
✅ Answer: The secant-line estimate is lower because the quadratic is concave down.

(C) The range of \(I\) is limited by the context:

  1. The number of cones sold cannot be negative: \(I(t) \ge 0\).
  2. Since \(a < 0\), the quadratic has a maximum. From the model, the vertex gives the maximum number of cones sold. The vertex \(t = -\frac{b}{2a} \approx 29.23\) yields \(I_{\text{max}} \approx 58.64\).

Thus, a reasonable range is \(0 \le I(t) \le 59\) (whole cones).
✅ Answer: The range is limited to non‑negative integers up to about 59 cones.

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