Quadratic Models on the SAT
Three things to look for
A quadratic model rises and then falls, or falls and then rises. On the SAT it is usually the height of a thrown object, a profit, or an area. Almost every question asks about one of three things: the starting value, the vertex, or the zeros.
For $h(t) = at^2 + bt + c$, the starting value is $h(0) = c$, the vertex is at $t = -\dfrac{b}{2a}$, and the zeros are where $h(t) = 0$. Decide which of the three the question wants before you calculate anything.
Quadratic models often appear among the harder questions in the second module, because one formula can carry several different questions. The algebra is the same each time. What students find hard is deciding which part of the graph the words are asking about.
$h(t) = -16t^2 + 64t + 5$. The starting height is $h(0) = 5$, the highest point is the vertex $(2, 69)$, and the ball lands at the positive zero, about $t = 4.1$.
Matching the words to the graph
| The question asks | Find this |
|---|---|
| "Initial height", "launched from", "fixed cost" | The value at $0$: the constant $c$. |
| "Maximum height", "greatest profit" | The output at the vertex. Find $t = -\dfrac{b}{2a}$, then put it back into the model. |
| "When does it reach its highest point?" | The input at the vertex, $t = -\dfrac{b}{2a}$. |
| "When does it hit the ground?" | The positive zero: solve $h(t) = 0$. |
| Vertex form $a(t - h)^2 + k$ | The vertex is $(h, k)$. You can read it without any calculation. |
| "For how long is it above $16.7$ meters?" | Solve $h(t) = 16.7$, then subtract the two times. |
Four examples
Example 1: the three key values
A ball's height in feet after $t$ seconds is $h(t) = -16t^2 + 64t + 5$. Find the starting height and the maximum height.
- Starting height: $h(0) = 5$ feet.
- Vertex: $t = -\dfrac{64}{2(-16)} = 2$ seconds.
- Maximum height: $h(2) = -16(4) + 128 + 5 = 69$ feet. The answer $2$ is when, and $69$ is how high. Check which one the question is asking for.
Example 2: when it hits the ground
A rock's height in meters is $h(t) = -5t^2 + 20t + 25$. When does it hit the ground?
- Set $h(t) = 0$ and divide by $-5$: $t^2 - 4t - 5 = 0$.
- Factor: $(t - 5)(t + 1) = 0$, so $t = 5$ or $t = -1$.
- Time cannot be negative, so the rock lands after $5$ seconds.
Example 3: a maximum in a money problem
A store's revenue at price $p$ dollars is $R(p) = p(200 - 4p)$. What price gives the most revenue?
- The zeros are $p = 0$ and $p = 50$.
- A parabola is symmetric, so the vertex is halfway between the zeros: $p = 25$.
- The maximum revenue is $R(25) = 25 \cdot 100 = 2{,}500$ dollars.
Example 4: building the model from three values
A quadratic $f(x) = ax^2 + bx + c$ has $f(0) = 10$, $f(1) = 13$ and $f(2) = 12$. Find $f(3)$.
- $f(0) = c = 10$.
- $f(1)$: $a + b + 10 = 13$, so $a + b = 3$. $f(2)$: $4a + 2b + 10 = 12$, so $2a + b = 1$.
- Subtract: $a = -2$, then $b = 5$. So $f(3) = -18 + 15 + 10 = 7$.
Mistakes to avoid
- Giving the time when the question asks for the height. The vertex has two numbers. A question that asks "when" wants the first number, and one that asks "how high" wants the second.
- Keeping a negative zero. In a time or length problem, throw away the negative solution.
- Sign slips in $-\dfrac{b}{2a}$. With $a = -16$ and $b = 64$, the vertex is at $t = 2$, not $-2$.
- Reading vertex form backward. In $-16(t - 3)^2 + 150$, the vertex is $(3, 150)$, not $(-3, 150)$, and the starting height is $h(0)$, not $150$.
- Calling $c$ the maximum. The constant is only the value at $0$. The maximum is at the vertex.
Practice questions
Underline what each question asks for, the time or the height, before you start.
Easy
The height, in feet, of a ball $t$ seconds after it is thrown is $h(t) = -16t^2 + 64t + 5$. From what height was the ball thrown?
- A$5$ feet
- B$64$ feet
- C$-16$ feet
- D$53$ feet
Show solution
Answer: A. The ball is thrown at $t = 0$, and $h(0) = 5$. Choice D is the height after $1$ second.
EasyStudent-produced response
The height, in feet, of a ball $t$ seconds after it is thrown is $h(t) = -16t^2 + 64t + 5$. What is the height of the ball, in feet, after $2$ seconds?
Enter your answer.
Show solution
Answer: $69$. $h(2) = -16(4) + 64(2) + 5 = -64 + 128 + 5 = 69$.
Medium
The height, in feet, of a ball $t$ seconds after it is thrown is $h(t) = -16t^2 + 64t + 5$. What is the maximum height the ball reaches?
- A$2$ feet
- B$53$ feet
- C$64$ feet
- D$69$ feet
Show solution
Answer: D. The vertex is at $t = -\dfrac{64}{2(-16)} = 2$. The maximum height is $h(2) = 69$ feet. Choice A is the time of the maximum, not the height.
MediumStudent-produced response
The height, in meters, of a rock $t$ seconds after it is thrown is $h(t) = -5t^2 + 20t + 25$. How many seconds after it is thrown does the rock hit the ground?
Enter your answer.
Show solution
Answer: $5$. Set $-5t^2 + 20t + 25 = 0$ and divide by $-5$: $t^2 - 4t - 5 = 0$, so $(t - 5)(t + 1) = 0$. The positive solution is $t = 5$.
Medium
A store's revenue, in dollars, when it charges $p$ dollars for an item is $R(p) = p(200 - 4p)$. What is the maximum revenue?
- A\$2,500
- B\$25
- C\$50
- D\$10,000
Show solution
Answer: A. The zeros are $p = 0$ and $p = 50$, so the vertex is at $p = 25$. $R(25) = 25(200 - 100) = 2{,}500$. Choice B is the best price, not the revenue.
MediumStudent-produced response
A rectangle has a perimeter of $40$ meters. Its area, in square meters, is $A(w) = w(20 - w)$, where $w$ is its width in meters. What is the greatest possible area of the rectangle?
Enter your answer.
Show solution
Answer: $100$. The zeros are $w = 0$ and $w = 20$, so the vertex is at $w = 10$. $A(10) = 10 \cdot 10 = 100$ square meters. The largest rectangle is a square.
Hard
The height, in feet, of a ball $t$ seconds after it is launched is $h(t) = -16(t - 3)^2 + 150$. Which statement is true?
- AThe ball reaches a maximum height of $3$ feet after $150$ seconds.
- BThe ball is launched from a height of $150$ feet.
- CThe ball reaches a maximum height of $150$ feet after $3$ seconds.
- DThe ball hits the ground after $3$ seconds.
Show solution
Answer: C. In vertex form $a(t - h)^2 + k$ the vertex is $(h, k) = (3, 150)$. Since $a$ is negative, $150$ is the maximum. The launch height is $h(0) = -144 + 150 = 6$ feet, so choice B is false.
HardStudent-produced response
The function $f(x) = ax^2 + bx + c$ satisfies $f(0) = 10$, $f(1) = 13$ and $f(2) = 12$. What is the value of $f(3)$?
Enter your answer.
Show solution
Answer: $7$. $c = 10$. Then $a + b = 3$ and $4a + 2b = 2$, so $2a + b = 1$. Subtracting gives $a = -2$ and $b = 5$. $f(3) = -2(9) + 5(3) + 10 = 7$.
Hard
A company's profit, in dollars, from selling $x$ units is $P(x) = -2x^2 + 120x - 1{,}000$. What is the maximum profit the company can make?
- A\$30
- B\$1,000
- C\$800
- D\$3,600
Show solution
Answer: C. The vertex is at $x = -\dfrac{120}{2(-2)} = 30$ units. $P(30) = -2(900) + 3{,}600 - 1{,}000 = 800$. Choice A is the number of units, not the profit.
HardStudent-produced response
The height, in meters, of a ball $t$ seconds after it is thrown is $h(t) = -4.9t^2 + 19.6t + 2$. For how many seconds is the ball higher than $16.7$ meters?
Enter your answer.
Show solution
Answer: $2$. Set $-4.9t^2 + 19.6t + 2 = 16.7$, so $-4.9t^2 + 19.6t - 14.7 = 0$. Divide by $-4.9$: $t^2 - 4t + 3 = 0$, so $t = 1$ or $t = 3$. The ball is above $16.7$ meters from $t = 1$ to $t = 3$, which is $2$ seconds.
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