Interpreting Quadratic Models in Context
Projectile and revenue problems are quadratics. The vertex gives the maximum (or minimum); the zeros give where the output is zero.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
The vertex's $x$-coordinate is when the max/min happens; its $y$-coordinate is the max/min value. Don't confuse the two, and don't mistake the vertex for a zero.
How to solve one, step by step
Example: $h(t) = -16t^2 + 32t$ gives height in feet.
- Vertex time: $t = -\frac{32}{2(-16)} = 1$.
- Max height: $h(1) = -16 + 32 = 16$ feet.
The mistakes that cost points
- Mistaking the vertex for a zero. The vertex is the peak, not where the height is $0$.
- Confusing the max value with the time of the max. The $y$ of the vertex is the value; the $x$ is when it occurs.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
A ball is launched upward and its height in feet after $t$ seconds is modeled by $h(t) = -16t^2 + 64t + k$, where $k$ is the initial height. The ball reaches its maximum height at $t = 2$ seconds. Which of the following best describes what occurs at $t = 2$?
- AThe ball is at its highest point — the vertex of the parabola.
- BThe parabola opens upward, so $t = 2$ is the minimum.
- CThe maximum height equals 2 feet.
- DThe ball hits the ground at $t = 2$.
Show solution
Answer: A, The ball is at its highest point — the vertex of the parabola.. The maximum of a downward-opening parabola occurs at the vertex. For $h(t) = -16t^2 + 64t + k$, the vertex is at $t = -\frac{64}{2(-16)} = 2$. At $t = 2$, the ball is at its maximum height — not at the ground, not at $h = 0$.
Medium
A store's daily revenue is modeled by $R(p) = -2p^2 + kp$, where $p$ is the price in dollars. The revenue is maximized when $p = \$8$. What is the value of $k$?
- A$k = -32$
- B$k = 0$
- C$k = 32$
- D$k = 8$
Show solution
Answer: C, $k = 32$. The vertex of $R(p) = -2p^2 + kp$ is at $p = \frac{k}{4}$. Setting $\frac{k}{4} = 8 \Rightarrow k = 32$.
Hard
A football's height above the ground (in feet) during a kick is modeled by $H(t) = -16t^2 + kt - 5 + c$, where $c$ shifts the model. Suppose instead $H(t) = -16t^2 + 48t + k$. The ball lands (returns to height 0) at $t = 3$ seconds. What is the value of $k$?
- A$k = 0$
- B$k = 48$
- C$k = 144$
- D$k = 3$
Show solution
Answer: A, $k = 0$. $H(3) = 0$: $-16(9) + 48(3) + k = -144 + 144 + k = k = 0$. The initial height is $k = 0$, meaning the ball was kicked from ground level.
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