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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