AP Precalculus -1.3 Rates of Change in Linear and Quadratic Functions- MCQ Exam Style Questions - Effective Fall 2023
AP Precalculus -1.3 Rates of Change in Linear and Quadratic Functions- MCQ Exam Style Questions – Effective Fall 2023
AP Precalculus -1.3 Rates of Change in Linear and Quadratic Functions- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.
▶️ Answer/Explanation
For a piecewise-linear function with two linear segments, the average rate of change should be roughly constant in each segment but different between segments.
- \( f \): rates ~constant ⇒ one linear function, not two segments.
- \( g \): rates increase steadily ⇒ curved, not piecewise-linear.
- \( h \): rates ~2.1 for intervals 1–3 (0<x<3), then ~4.2 for intervals 4–7 (3<x<7) ⇒ two different constant slopes ⇒ piecewise-linear with two segments.
- \( k \): rates increase then decrease ⇒ not constant in two blocks.
✅ Answer: (C)
▶️ Answer/Explanation
Average rate of change from \( a \) to \( a+1 \) is \( 2a + 1 \), which is linear in \( a \).
If \( f \) were linear, the average rate of change would be constant (not depend on \( a \)), so (A) and (C) are false.
If \( f \) is quadratic, then the average rate of change over a 1-unit interval is indeed linear in \( a \) (this can be derived). Here \( 2a + 1 \) increases as \( a \) increases, at a constant rate (slope 2 with respect to \( a \)), meaning \( f \) is concave up (opens upward).
Thus the correct reasoning: the average rate of change is increasing at a constant rate ⇒ \( f \) could be a parabola opening up.
✅ Answer: (D)
▶️ Answer/Explanation
Using points (3, -2) and (6, 2): slope \( m = \frac{2 – (-2)}{6 – 3} = \frac{4}{3} \).
Equation: \( f(x) – (-2) = \frac{4}{3}(x – 3) \) ⇒ \( f(x) = \frac{4}{3}x – 4 – 2 = \frac{4}{3}x – 6 \).
Check with (-3, -10): \( \frac{4}{3}(-3) – 6 = -4 – 6 = -10 \) ✓.
Now \( f(13) = \frac{4}{3}(13) – 6 = \frac{52}{3} – \frac{18}{3} = \frac{34}{3} \).
✅ Answer: (D)
▶️ Answer/Explanation
For a quadratic function \( p(x) = ax^2 + bx + c \), the average rate of change over equal-length intervals changes linearly.
Given:
Average rate on \( [0,2] \) is \( -4 \).
Average rate on \( [2,4] \) is \( -1 \).
The change from one interval to the next is constant: \( -1 – (-4) = 3 \).
Thus, for each subsequent interval of length 2, the average rate increases by 3.
Sequence of average rates:
\( [0,2]: -4 \)
\( [2,4]: -1 \)
\( [4,6]: -1 + 3 = 2 \)
\( [6,8]: 2 + 3 = 5 \)
✅ Answer: (C) 5
▶️ Answer/Explanation
Calculate the average rate of change (speed) over each time interval:
From \( t=1 \) to \( t=3 \): \( \frac{9-1}{3-1} = \frac{8}{2} = 4 \) m/s
From \( t=3 \) to \( t=6 \): \( \frac{21-9}{6-3} = \frac{12}{3} = 4 \) m/s
From \( t=6 \) to \( t=11 \): \( \frac{41-21}{11-6} = \frac{20}{5} = 4 \) m/s
The speed is constant at 4 m/s, so the acceleration (rate of change of speed) is 0 m/s².
Therefore, the rate of change of the rates of change is \( 0 \) m/s², and the object moves with constant speed (neither speeding up nor slowing down).
✅ Answer: (B)
▶️ Answer/Explanation
\(
\begin{array}{c|cccc}
x & -2 & -1 & 0 & 1 \\
\hline
f(x) & 5 & 2 & 1 & 2
\end{array}
\)
Step 1: Rates of change over equal intervals
All \(x\)-intervals have length \(1\).
\([-2,-1] : \quad 2 – 5 = -3 \)
\([-1,0]: \quad 1 – 2 = -1 \)
\([0,1]: \quad 2 – 1 = 1\)
So the rates of change are
\(
-3,\,-1,\,1.
\)
Step 2: Identify the pattern
The rates of change are not constant, so \(f\) is not linear.
Now check whether the rates follow a linear rule.
Label each interval by its left endpoint \(x=-2,-1,0\):
\(
2x+1 =
\begin{cases}
2(-2)+1=-3, \\
2(-1)+1=-1, \\
2(0)+1=1.
\end{cases}
\)
This matches the observed rates of change.
Step 3: Conclusion
Since the first differences follow a linear pattern, \(f\) could be a quadratic function.
\(
\boxed{\text{Correct answer: (D)}}
\)
▶️ Answer/Explanation
We have:
\( g(x) = f(x+1) – f(x) \) → average rate of change of \( f \) over interval \([x, x+1]\).
\( h(x) = g(x+1) – g(x) \) → change in \( g \) from \( x \) to \( x+1 \).
Given \( h(x) = -6 \) for all \( x \).
Thus \( g(x+1) – g(x) = -6 \) ⇒ \( g \) is decreasing linearly with slope \(-6\).
But \( g \) itself is the rate of change of \( f \) (over 1-unit intervals).
Since \( g \) is decreasing, the rate of change of \( f \) is decreasing ⇒ the graph of \( f \) is concave down.
Also, \( g \) decreasing does **not** mean \( g \) is negative, only that it’s getting smaller. So \( f \) could still have positive slope if \( g > 0 \).
Thus: \( h \) constant negative ⇒ \( g \) decreasing ⇒ \( f \) concave down.
✅ Answer: (D)