SAT Nonlinear functions Practice Question 13 (Medium)

Math · Advanced MathMedium

The graph of a quadratic function in the $xy$ -plane has $x$ -intercepts at $(-1, 0)$ and $(5, 0)$ and a $y$ -intercept at $(0, -10)$ . Which equation defines the function?

$y = 2(x + 1)(x - 5)$

$y = (x + 1)(x - 5)$

$y = 2(x - 1)(x + 5)$

$y = -2(x + 1)(x - 5)$

Show answer and explanation

Answer: A

The zeros give $y = a(x + 1)(x - 5)$ . Substituting $(0, -10)$ : $-10 = a(1)(-5)$ , so $a = 2$ . The equation is $y = 2(x + 1)(x - 5)$ .

Choice B has the correct zeros but a $y$ -intercept of $-5$ , not $-10$ .

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