Interpreting Roots and Zeros in Context
A zero is an input where the output equals zero — in a context, often the moment something hits the ground or breaks even.
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 zeros are the $x$-values that make the function $0$. In context, read what $x$ and the output stand for: a zero is an $x$-value, not a $y$-value.
How to solve one, step by step
Example: $h(t) = -(t - 2)(t - 5)$ gives height over time.
- The zeros are $t = 2$ and $t = 5$.
- These are the times when the height is $0$.
The mistakes that cost points
- Reporting the output at the zero. A zero is the $x$-value; the output there is $0$.
- Confusing how many roots with their values. "Two zeros" is a count; $t = 2, 5$ are the values.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
The function $h(t) = -t^2 + 6t$ models the height, in meters, of a projectile $t$ seconds after launch. For how many seconds is the projectile above the ground?
- A$2$
- B$3$
- C$6$
- D$9$
Show solution
Answer: C, $6$. Find where $h(t) = 0$: $-t^2 + 6t = 0 \Rightarrow -t(t - 6) = 0 \Rightarrow t = 0$ or $t = 6$. The projectile is above the ground between $t = 0$ and $t = 6$, so for $6$ seconds.
Medium
For which integer value of $k$ will the parabola $22(y+k) = (x+1)^2$ have integer $x$-intercepts?
- A$k = 484$
- B$k = 1$
- CThere is no such value of $k$
- D$k = 22$
Show solution
Answer: D, $k = 22$. Set $y = 0$: $22(0+k) = (x+1)^2$, which simplifies to $22k = (x+1)^2$. For $x$ to be an integer, $22k$ must be a perfect square. The smallest positive integer $k$ that makes $22k$ a perfect square is $k = 22$, giving $22 \cdot 22 = 484 = 22^2$. Then $(x+1)^2 = 484$, so $x + 1 = \pm 22$, giving $x = 21$ or $x = -23$. Both are integers.
Hard
The function $f(x) = 2x^2 - kx - 9$ has a zero at $x = -\frac{1}{2}$. What is the value of $k$?
- A$-\frac{1}{2}$
- B$17$
- C$\frac{17}{2}$
- D$-17$
Show solution
Answer: B, $17$. Substitute $x = -\frac{1}{2}$ into $2x^2 - kx - 9 = 0$: $2\left(\frac{1}{4}\right) - k\left(-\frac{1}{2}\right) - 9 = 0 \Rightarrow \frac{1}{2} + \frac{k}{2} - 9 = 0 \Rightarrow \frac{k}{2} = \frac{17}{2} \Rightarrow k = 17$.
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