Question

Let $f(x)$ be a quadratic polynomial in $x$ such that $f(x) \ge 0$ for all real numbers $x$. If $f(2) = 0$ and $f(4) = 6$, then $f(-2)$ is equal to

Accepted Answer

C. Since $f(x) \ge 0$ for all $x$ and $f(2) = 0$, $x = 2$ must be a minimum of the quadratic polynomial, so $x = 2$ is a double root. Therefore, $f(x) = a(x - 2)^2$ for some $a > 0$. Given $f(4) = 6$: $a(4 - 2)^2 = 6 \implies 4a = 6 \implies a = \frac{3}{2}$. Thus $f(x) = \frac{3}{2}(x - 2)^2$. $f(-2) = \frac{3}{2}(-2 - 2)^2 = \frac{3}{2}(16) = 24$.

Quadratic Polynomial Properties

Explanation

Practise this under exam conditions