# How do you find the vertex of a parabola g(x) = x^2 - 4x + 2?

Jun 28, 2015

I found (coordinates of the vertex):
${x}_{v} = 2$
${y}_{v} = - 2$

#### Explanation:

You have two main ways to find the coordinates of the vertex:
1] your parabola is in the form $a {x}^{2} + b x + c$
Where:
$a = 1$
$b = - 4$
$c = 2$
The coordinates of the vertex are then:
$\textcolor{red}{{x}_{v} = - \frac{b}{2 a}} = - \frac{- 4}{2 \cdot 1} = 2$
$\textcolor{red}{{y}_{v} = - \frac{\Delta}{4 a}} = - \frac{{b}^{2} - 4 a c}{4 a} = - \frac{16 - 8}{4} = - 2$

2] Use the derivative.
At the vertex the derivative of your function must be ZERO;
So:
derivative $g ' \left(x\right) = 2 x - 4$ set it equal to zero and solve for $x$:
$2 x - 4 = 0$
$x = \frac{4}{2} = 2 = {x}_{v}$
use this value into your original function to find ${y}_{v}$:
$g \left(2\right) = 4 - 8 + 2 = - 2 = {y}_{v}$