Question 231652
<font face="Garamond" size="+2">


I presume you mean *[tex \Large y\ =\ ax^2\ +\ bx\ +\ c]


First, factor an *[tex \Large a] out of the first two terms on the right.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ a\left(x^2\ +\ \frac{b}{a}x\right)\ + c]


Complete the square:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ a\left(x^2\ +\ \frac{b}{a}x\ + \frac{b^2}{4a^2}\right)\ -\ a\left(\frac{b^2}{4a^2}\right) + c]


Factor the square:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ a\left(x\ - \left(-\frac{b}{2a}\right)\right)\ -\ \frac{b^2}{4a} + c]


Then, by inspection:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ a]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h\ =\ -\frac{b}{2a}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ =\ -\ \frac{b^2}{4a} + c]






John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>