# How do you find the vertex for y=x^2-10x+20?

Apr 7, 2016

color(green)("vertex: " (5,-5)

#### Explanation:

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

Converting $y = {x}^{2} - 10 x + 20$ into vertex form:

Complete the square:
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} - 10 x + {5}^{2} + 20 - {5}^{2}$

Simplify into vertex form
$\textcolor{w h i t e}{\text{XXX}} y = 1 {\left(x - 5\right)}^{2} - 5$
with vertex at $\left(5 , - 5\right)$
graph{x^2-10x+20 [-5.13, 14.87, -7.32, 2.68]}