▶️ Answer/Explanation
A point of inflection occurs where the graph changes concavity (from concave up to concave down, or vice-versa).
Starting from the left, the graph begins concave down as it rises toward the first local maximum.
It switches to concave up to form the first local minimum near the $y$-axis (Point $1$).
It switches back to concave down to form the second local maximum (Point $2$).
It then switches to concave up to form the deep local minimum on the right (Point $3$).
Finally, it switches to concave down very briefly or stays concave up? Looking closely, the final upward stroke suggests a $4^{th}$ change in curvature (Point $4$).
Counting the transitions between the peaks and valleys, there are four distinct inflection points.
Correct Option: (C)
▶️ Answer/Explanation
The table shows zeros at $x = 1$ and $x = 5$, where $f(1) = 0$ and $f(5) = 0$.
Between these zeros, at $x = 3$, the function is positive ($f(3) = 4$).
For $x < 1$, specifically at $x = -1$, the function is negative ($f(-1) = -36$).
For $x > 5$, specifically at $x = 7$, the function is positive ($f(7) = 12$).
Since $f(x)$ must cross or touch the x-axis only at $x=1$ and $x=5$, the sign change at $x=1$ (negative to positive) indicates it is not a maximum.
However, at $x = 5$, the function goes from positive ($f(3)=4$) to zero ($f(5)=0$) and back to positive ($f(7)=12$).
This means $f(x)$ touches the x-axis at $(5, 0)$ without crossing it, staying $\ge 0$ in that interval.
Therefore, $(5, 0)$ must be a local minimum.
▶️ Answer/Explanation
The concavity of a function $f$ is determined by the behavior of its rate of change, $f’$.
A graph is concave down on an interval if its rate of change is decreasing.
In the interval $0 < x < 1$, the rate of change is increasing, so $f$ is concave up.
In the interval $1 < x < 2$, the rate of change is constant, so there is no concavity.
In the interval $2 < x < 3$, the table explicitly states the rate of change is decreasing.
In the interval $3 < x < 4$, the rate of change is constant, so there is no concavity.
Therefore, the graph of $f$ is concave down only on the interval $2 < x < 3$.
Correct Option: (C)
▶️ Answer/Explanation
For a linear function, the rate of change remains constant.
A piecewise-linear function with two segments must show two distinct constant rates.
Function $f$ has a nearly constant rate of $\approx 2.1$, suggesting a single linear model.
Function $g$ shows a constantly increasing rate, suggesting a quadratic model.
Function $h$ stays at $\approx 2.1$ for $0 < x < 3$ and then jumps to $\approx 4.1$ for $3 < x < 7$.
This clear shift between two stable values identifies two linear segments with different slopes.
Function $k$ increases and then decreases, which is characteristic of a single nonlinear curve.
Therefore, $h$ is the best fit for a piecewise-linear model with two segments.
▶️ Answer/Explanation
The “rate of change” of $g$ refers to the first derivative, $g'(x)$.
If $g'(x)$ is increasing for $x < 2$, then the second derivative $g”(x) > 0$.
A positive second derivative ($g”(x) > 0$) means the graph is concave up.
If $g'(x)$ is decreasing for $x > 2$, then the second derivative $g”(x) < 0$.
A negative second derivative ($g”(x) < 0$) means the graph is concave down.
Since the concavity changes at $x = 2$, there is a point of inflection at $x = 2$.
Therefore, the correct choice is (D).
▶️ Answer/Explanation
The total change in height for Drone A is $17 + (-4) + 11 + (-5) + (-3) = 16$ feet.
The total change in height for Drone B is $5 + 3 + 3 + 2 + 3 = 16$ feet.
Average rate of change is calculated as $\frac{\text{Total Change in Height}}{\text{Total Time}}$.
For Drone A, the average rate of change is $\frac{16}{30}$ feet per second.
For Drone B, the average rate of change is $\frac{16}{30}$ feet per second.
Since both total changes and total times are identical, the average rates are equal.
Therefore, the correct choice is (A).
▶️ Answer/Explanation
For a quadratic function, the average rate of change over equal-length adjacent intervals forms an arithmetic progression.
The length of the given intervals is constant at $\Delta x = 2$.
The first interval $[0, 2]$ has an average rate of change of $-4$.
The second interval $[2, 4]$ has an average rate of change of $-1$.
The common difference between these rates is $(-1) – (-4) = 3$.
The third interval $[4, 6]$ would have a rate of $(-1) + 3 = 2$.
The fourth interval $[6, 8]$ would have a rate of $2 + 3 = 5$.
Therefore, the average rate of change on $6 \leq x \leq 8$ is $5$.
▶️ Answer/Explanation
The correct option is (D).
First, calculate the average rates of change for each interval of width $\Delta x = 1$:
For $[-2, -1]$, the rate is $\frac{2 – 5}{-1 – (-2)} = -3$.
For $[-1, 0]$, the rate is $\frac{1 – 2}{0 – (-1)} = -1$.
For $[0, 1]$, the rate is $\frac{2 – 1}{1 – 0} = 1$.
Since the rates of change are not constant, the function cannot be linear.
The rates of change $\{-3, -1, 1\}$ follow a linear pattern described by $y = 2x + 1$ (where $x$ is the left endpoint).
Because the first differences are linear, the function $f$ could be quadratic.
▶️ Answer/Explanation
The original function is $f(x) = x^2 + 2$.
The transformed function is $g(x) = \frac{x^2}{4} + 2$, which can be rewritten as $g(x) = (\frac{x}{2})^2 + 2$.
Comparing the two, we see that $g(x) = f(\frac{x}{2})$.
A transformation of the form $f(\frac{x}{k})$ represents a horizontal dilation by a factor of $k$.
In this case, $k = 2$, which means the graph is stretched horizontally.
Therefore, the transformation is a horizontal dilation by a factor of $2$.
The correct option is (C).
▶️ Answer/Explanation
The domain of $g$ is found by setting the argument $x + 3$ within the domain of $f$: $-2 \le x + 3 \le 2$.
Subtracting $3$ from all sides gives the new domain: $-5 \le x \le -1$, or $[-5, -1]$.
For the range, the original outputs $f(x)$ are in the interval $[1, 5]$.
Multiplying the range by $-2$ scales and reflects it: $[-10, -2]$.
Adding $4$ to this interval shifts it to $[-10 + 4, -2 + 4] = [-6, 2]$.
Therefore, the domain is $[-5, -1]$ and the range is $[-6, 2]$.
The correct option is (A).
▶️ Answer/Explanation
Observe $f(x)$ peaks at $f(3) = 9$ within the first period $x \in [0, 6]$.
Observe $g(x)$ peaks at $g(1) = 2$ and $g(2) = 2$ within its first period.
The maximum value of $f$ is $9$ and the maximum value of $g$ is $2$, implying a vertical dilation by $\frac{2}{9}$.
However, looking at the pattern, $f$ returns to $0$ at $x=6$, while $g$ returns to $0$ at $x=3$.
This indicates the period of $g$ is half the period of $f$, meaning a horizontal dilation by a factor of $\frac{1}{2}$.
Comparing $f(1)=1$ to $g(x)$ values, a vertical dilation by factor $\frac{1}{3}$ relates $f(3)=9$ to $g$ peaks if looking at specific points.
By analyzing the transformation $g(x) = \frac{1}{3} f(2x)$, we see $g(1) = \frac{1}{3}f(2) = \frac{4}{3}$ (approx 2 in discrete table) and $g(1.5) = \frac{9}{3}=3$.
Based on the provided options, (B) is the closest fit for the observed period change and amplitude reduction.
▶️ Answer/Explanation
Boyle’s Law implies an inverse relationship: $P \cdot V = k$, where $k$ is a constant.
Using the first data point from the table: $P = 137.500$ and $V = 12$.
Calculate the constant: $k = 137.500 \times 12 = 1650$.
Verify with another point: $103.125 \times 16 = 1650$.
Rearranging the equation $P \cdot V = 1650$ to solve for $V$ gives $V(P) = \frac{1650}{P}$.
Therefore, the correct model matching the data is option (D).
▶️ Answer/Explanation
The initial velocity is $v_0 = 4.4$ m/s (positive), so the ball first moves upward.
The time to reach maximum height is $t = \frac{-b}{2a} = \frac{-4.4}{2(-4.9)} \approx 0.449$ seconds.
The maximum height is $h(0.449) = -4.9(0.449)^2 + 4.4(0.449) + 15.24 \approx 16.228$ meters.
To find when it hits the ground, set $h(t) = 0$ and solve $-4.9t^2 + 4.4t + 15.24 = 0$ using the quadratic formula.
The total time $t \approx 2.269$ seconds is the time from $t=0$ until impact.
The time from maximum height to the ground is $2.269 – 0.449 = 1.820$ seconds.
Therefore, Option (C) correctly describes the upward motion to $16.228$ m and the subsequent $1.820$ s fall.
▶️ Answer/Explanation
The relationship is given as $I \propto \frac{1}{d^2}$, which implies $I \cdot d^2 = k$.
Using the initial values, calculate the constant: $k = 26.5 \cdot 4^2 = 26.5 \cdot 16 = 424$.
Set up the equation for the new distance: $I_{new} \cdot (3.3)^2 = 424$.
Calculate the new distance squared: $(3.3)^2 = 10.89$.
Solve for the new intensity: $I_{new} = \frac{424}{10.89}$.
Perform the division: $I_{new} \approx 38.9348$.
The result rounds to $38.935$, which matches option (D).
▶️ Answer/Explanation
a. Graphing the function
Using a graphing utility for \(P(x) = -0.0013x^3 + 0.3507x^2 – 0.4591x – 421.888\) on the interval \([0, 200]\) reveals a cubic curve shape.
The graph starts with a slight dip to a local minimum near \(x=0\), then rises steeply to a local maximum near \(x=180\), before falling again.
b. Average rate of change between extrema
First, find the derivative: \(P'(x) = -0.0039x^2 + 0.7014x – 0.4591\).
Set \(P'(x) = 0\) and use the quadratic formula to find the extrema: \(x \approx 0.66\) (local min) and \(x \approx 179.19\) (local max).
Calculate the profit at these points: \(P(0.66) \approx -422.04\) and \(P(179.19) \approx 3276.57\).
The average rate of change is: \(\frac{3276.57 – (-422.04)}{179.19 – 0.66} = \frac{3698.61}{178.53} \approx 20.72\).
c. Equation of the secant line
The slope \(m\) was found in part (b) to be approximately \(20.72\).
Using the point-slope form \(y – y_1 = m(x – x_1)\) with the minimum point \((0.66, -422.04)\):
\(y – (-422.04) = 20.72(x – 0.66)\)
\(y = 20.72x – 13.68 – 422.04\)
The equation is approximately \(y = 20.72x – 435.72\).
d. Inflection point
Find the second derivative: \(P”(x) = -0.0078x + 0.7014\).
Set \(P”(x) = 0\) to find the change in concavity: \(0 = -0.0078x + 0.7014 \Rightarrow x = \frac{0.7014}{0.0078} \approx 89.92\).
Find the corresponding y-value: \(P(89.92) \approx 1427.27\).
The inflection point is approximately \((89.92, 1427.27)\).
e. Variation of rate of change
The rate of change is represented by the derivative, \(P'(x)\).
Before the inflection point (\(x < 89.92\)), the graph is concave up (\(P”(x) > 0\)), so the rate of change is increasing.
After the inflection point (\(x > 89.92\)), the graph is concave down (\(P”(x) < 0\)), so the rate of change is decreasing.
▶️ Answer/Explanation
To transform the graph, apply a horizontal translation left by \(\frac{1}{2}\) unit followed by a horizontal compression by a factor of \(\frac{2}{3}\).
Reasoning: Shifting left by \(\frac{1}{2}\) changes \(x \to x + \frac{1}{2}\), giving \(f(2(x+0.5)+1) = f(2x+2)\). Compressing by \(\frac{2}{3}\) changes \(x \to \frac{3}{2}x\), resulting in \(f(2(\frac{3}{2}x)+2) = f(3x+2)\).
First, apply a horizontal translation left by 1 unit to get \(y = f(x+1)\). Then, apply a horizontal compression by a factor of \(\frac{1}{2}\). This replaces \(x\) with \(2x\), resulting in \(y = f(2x+1)\).
First, apply a horizontal compression by a factor of \(\frac{1}{2}\) to get \(y = f(2x)\). Then, apply a horizontal translation left by \(\frac{1}{2}\) unit. This replaces \(x\) with \(x+\frac{1}{2}\), resulting in \(y = f(2(x+0.5)) = f(2x+1)\).