1957 AHSME Problems/Problem 40
Contents
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{align*} y|_{x=\tfrac 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\sqrt2 \end{align*} Because both of these results are irrational with one positive and one negtaive solution, we choose answer .
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 39 |
Followed by Problem 41 | |
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