1986 AHSME Problems/Problem 13

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Problem

A parabola $y = ax^{2} + bx + c$ has vertex $(4,2)$. If $(2,0)$ is on the parabola, then $abc$ equals

$\textbf{(A)}\ -12\qquad \textbf{(B)}\ -6\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 12$

Solution

Consider the quadratic in completed square form: it must be $y=a(x-4)^{2}+2$. Now substitute $x=2$ and $y=0$ to give $a=-\frac{1}{2}$. Now expanding gives $y=-\frac{1}{2}x^{2}+4x-6$, so the product is $-\frac{1}{2} \cdot 4 \cdot -6  = 3 \cdot 4 = 12$, which is $\boxed{E}$.

See also

1986 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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