What is the vertex of y=x^2-6x+15?

Jan 6, 2016

Vertex is at $\left(3 , 6\right)$

Explanation:

The general vertex form of a parabola is
$\textcolor{w h i t e}{\text{XXX}} y = m {\left(x - \textcolor{b l u e}{a}\right)}^{2} + \textcolor{b l u e}{b}$ for a parabola with vertex at $\left(\textcolor{b l u e}{a , b}\right)$

Converting $y = {x}^{2} - 6 x + 15$ into this form:

Complete the square:
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} - 6 x \textcolor{red}{+ {3}^{2}} + 15 \textcolor{red}{- {3}^{2}}$
Re-write as a squared binomial
$\textcolor{w h i t e}{\text{XXX}} y = {\left(x - \textcolor{b l u e}{3}\right)}^{2} + \textcolor{b l u e}{6}$
which is in the vertex form (with $m = 1$) for a parabola with vertex at $\left(\textcolor{b l u e}{3 , 6}\right)$