1957 AHSME Problems/Problem 10
Revision as of 20:28, 13 February 2021 by Justinlee2017 (talk | contribs) (Created page with "==Solution== This graph generates a parabola, since the degree of <math>x</math> is <math>2</math>. The <math>x-</math> coordinate of the vertex of a parabola given by <math>a...")
Solution
This graph generates a parabola, since the degree of 
is 
. The 
coordinate of the vertex of a parabola given by 
is at 
So, the vertex of this parabola is at
![\[x = \frac{-4}{2(2)} = \frac{-4}{4} = -1\]](http://latex.artofproblemsolving.com/b/1/1/b118824cc87f6384c130105d4db58c19436aa13b.png)
Since the coefficient of 
is positive, at 
, the parabola is at its minimum.
Substituting , we get
![\[y = 2(-1)^2 + 4(-1) + 3 \Rrightarrow y = 2 -4 + 3 \Rrightarrow y = 1\]](http://latex.artofproblemsolving.com/5/8/b/58b6f7ea40a816a643bfcad22d62a027320dc948.png)
So our answer is 
~JustinLee2017
