1965 AHSME Problems/Problem 9

Problem 9

The vertex of the parabola $y = x^2 - 8x + c$ will be a point on the $x$-axis if the value of $c$ is:

$\textbf{(A)}\ - 16 \qquad  \textbf{(B) }\ - 4 \qquad  \textbf{(C) }\ 4 \qquad  \textbf{(D) }\ 8 \qquad  \textbf{(E) }\ 16$


Notice that if the vertex of a parabola is on the x-axis, then the x-coordinate of the vertex must be a solution to the quadratic. Since the quadratic is strictly increasing on either side of the vertex, the solution must have double multiplicity, or the quadratic is a perfect square trinomial. This means that for the vertex of $y = x^2 - 8x + c$ to be on the x-axis, the trinomial must be a perfect square, and have discriminant of zero. So,

\begin{align*}     0 &= b^2-4ac\\     0 &= (-8)^2-4c\\     c &=  64\\     c &= 16\\ \end{align*}

Therefore $c=16$, and our answer is $\boxed{\textbf{(E)}}$

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