Difference between revisions of "1955 AHSME Problems/Problem 39"
(Wrote a Solution for this problem, since it didn't have one.) |
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<cmath>y = \frac{p^2}{4} - \frac{p^2}{2} + q</cmath> | <cmath>y = \frac{p^2}{4} - \frac{p^2}{2} + q</cmath> | ||
<cmath>0 = \frac{p^2}{4} - \frac{2p^2}{4} + q</cmath> | <cmath>0 = \frac{p^2}{4} - \frac{2p^2}{4} + q</cmath> | ||
| − | < | + | <cmath>0 = \frac{-p^2}{4} + q</cmath> |
<cmath>q = \frac{p^2}{4} = \boxed{B}</cmath> | <cmath>q = \frac{p^2}{4} = \boxed{B}</cmath> | ||
~JustinLee2017 | ~JustinLee2017 | ||
Revision as of 18:37, 12 February 2021
Solution
The least possible value of 
is given at the coordinate of the vertex. The 
- coordinate is given by
![\[\frac{-p}{2}\]](http://latex.artofproblemsolving.com/7/c/9/7c99e20f2c679f85c800b622be00bb81be6a184c.png)
. Plugging this into the quadratic, we get
![\[y = \frac{p^2}{4} - \frac{p^2}{2} + q\]](http://latex.artofproblemsolving.com/6/a/2/6a2afa567a387b10458f22e8fc82be8049ae82f9.png)
![\[0 = \frac{p^2}{4} - \frac{2p^2}{4} + q\]](http://latex.artofproblemsolving.com/a/7/5/a759167c387574d4a12f3635947c8284350ca532.png)
![\[0 = \frac{-p^2}{4} + q\]](http://latex.artofproblemsolving.com/9/e/5/9e5bca1b0f9c9eebe8b449b05c38e8bdd37c31f0.png)
![\[q = \frac{p^2}{4} = \boxed{B}\]](http://latex.artofproblemsolving.com/3/5/4/354bd18bcb66eab365a35361dd29404c5dccc2ed.png)
~JustinLee2017
