AP® Precalculus Free-Response Questions Past Paper 2025 Solution

i. Write three equations for the constants a, b, c.

We substitute each data point (t, D(t)) into the general quadratic equation D(t) = at² + bt + c.

1. For the point (0, 25):
   D(0) = a(0)² + b(0) + c = 25
   This simplifies to: c = 25
2. For the point (2, 30):
   D(2) = a(2)² + b(2) + c = 30
   This simplifies to: 4a + 2b + c = 30
3. For the point (4, 34):
   D(4) = a(4)² + b(4) + c = 34
   This simplifies to: 16a + 4b + c = 34

The three equations are:
1. c = 25
2. 4a + 2b + c = 30
3. 16a + 4b + c = 34

ii. Find the values for a, b, c.

We now solve this system of linear equations.

1. Use the value of c.
   From the first equation, we already know c = 25.
2. Substitute c into the other two equations.
   For equation 2: 4a + 2b + 25 = 30 ⇒ 4a + 2b = 5
   For equation 3: 16a + 4b + 25 = 34 ⇒ 16a + 4b = 9
3. Solve the new 2x2 system.
   We have:
   (Eq. A) 4a + 2b = 5
   (Eq. B) 16a + 4b = 9
   Multiply Eq. A by -2 to eliminate b: −2(4a + 2b) = −2(5) ⇒ −8a − 4b = −10.
   Add this new equation to Eq. B:
   (−8a − 4b) + (16a + 4b) = −10 + 9
   8a = −1 ⇒ a = −1/8 = −0.125
4. Find b.
   Substitute a = −0.125 back into Eq. A:
   4(−0.125) + 2b = 5
   −0.5 + 2b = 5
   2b = 5.5 ⇒ b = 2.75

The values for the constants are:
a = −0.125, b = 2.75, and c = 25.
The model is D(t) = −0.125t² + 2.75t + 25.

i. Find the average rate of change of the total number of plays from t = 0 to t = 4 months.

The average rate of change is calculated using the formula: (D(t₂) − D(t₁)) / (t₂ − t₁).

Using the points from the table, (t₁, D(t₁)) = (0, 25) and (t₂, D(t₂)) = (4, 34).

Average Rate of Change = (D(4) − D(0)) / (4 − 0) = (34 − 25) / 4

= 9 / 4 = 2.25

The average rate of change is 2.25 thousand plays per month.

ii. Use the average rate of change to estimate the total plays for t = 1.5 months.

We can form a linear approximation, A_t, based on the starting point D(0) and the average rate of change. The formula is A_t = D(0) + (Average Rate) × t.

A_1.5 = D(0) + (2.25) × 1.5

A_1.5 = 25 + 3.375 = 28.375

The estimated total number of plays at t = 1.5 months is 28.375 thousand plays.

iii. Explain why A_t < D(t) for all t where 0 < t < 4.

First, we show that A_1.5 < D(1.5). We calculated A_1.5 = 28.375. Now we find the actual value D(1.5) using our model from Part A.

D(1.5) = −0.125(1.5)² + 2.75(1.5) + 25
D(1.5) = −0.125(2.25) + 4.125 + 25
D(1.5) = −0.28125 + 4.125 + 25 = 28.84375

Indeed, 28.375 < 28.84375, so A_1.5 < D(1.5).

General Explanation:

The linear estimation A_t represents the secant line connecting the points (0, D(0)) and (4, D(4)) on the graph of D(t).

The function D(t) = −0.125t² + 2.75t + 25 is a quadratic with a negative leading coefficient (a = −0.125 < 0). This means the graph of D(t) is a parabola that opens downward, making it concave down everywhere.

For any concave down function, the graph of the function lies above its secant lines. Since A_t is the secant line on the interval from t=0 to t=4, the curve of D(t) must be above the line A_t for all values of t between 0 and 4. Therefore, A_t < D(t) for 0 < t < 4.

Explain how the maximum of the function D can be used to determine a boundary for the domain.

Real-World Context: The function D(t) represents the total number of plays since the song's release. A total, or cumulative, count cannot decrease over time. Therefore, for the model to be physically realistic, the function D(t) must be non-decreasing.

1. Analyze the Model's Behavior: Our function D(t) = −0.125t² + 2.75t + 25 is a downward-opening parabola, which means it increases to a maximum value and then decreases.
2. Find the Maximum: The maximum of the parabola occurs at its vertex. The t-coordinate of the vertex is given by the formula t = −b / (2a).
   t_vertex = −(2.75) / (2 × −0.125)
   t_vertex = −2.75 / −0.25 = 11
3. Interpret the Maximum: The vertex is at t = 11 months. This means the model predicts that the total number of plays increases for the first 11 months and then begins to decrease for t > 11.
4. Apply the Restriction: Since the total number of plays cannot decrease, the model is only valid on the interval where it is non-decreasing. This corresponds to the part of the parabola from the start (t=0) up to its vertex.

The maximum of the function occurs at t = 11 months. After this time, the model predicts a decrease in total plays, which contradicts the real-world scenario. Therefore, the time of the maximum, t = 11, serves as a realistic upper boundary for the domain of the model D. A reasonable domain for this model would be 0 ≤ t ≤ 11.