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:

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

Solution

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 $\boxed{\textbf{(E)}}$ .

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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1965 AHSME Problems/Problem 9

Problem 9

Solution

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