1957 AHSME Problems/Problem 40

Problem

If the parabola has its vertex on the -axis, then must be:

Solution 1

Note that if has its vertex on the -axis, then will have its vertex on the x-axis as well. To find the location of the vertex of the parabola, we desire to put it in vertex form, where , and is the location of the vertex. However, we know that , because the vertex is on the -axis. Thus, we know that must be the square of a linear term. Thus, , which are both irrational. Thus, our answer is .

Solution 2

We know that if a parabola is given by , then the -value of the vertex is (this fact can be proven with the quadratic formula and also derivatives). Because, in this case, , . Thus, at , the parabola should have a -value of . Therefore, we have the following equation that we can solve for : $$\begin{array}{rl}y{|}_{x=\frac{b}{2}}=-(\frac{b}{2}{)}^2-b\cdot (\frac{b}{2})-8& =0\\ -\frac{b^2}{4}+\frac{b^2}{2}& =8\\ \frac{b^2}{4}& =8\\ b^2& =32\\ b& =\pm \sqrt{32}=\pm 4\sqrt{2}\end{array}$$ Because both of these results are irrational with one positive and one negtaive solution, we choose answer .

See Also

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