# How do you find the coordinates for the vertex of f(x)=2x^2-8x+1?

Nov 13, 2016

The vertex is at $\left(2 , - 7\right)$.

#### Explanation:

$f \left(x\right) = 2 {x}^{2} - 8 x + 1$

We have to put the equation into vertex form by completing the square.
$f \left(x\right) = 2 \left[{x}^{2} - 4 x + \frac{1}{2}\right]$

$f \left(x\right) = 2 \left[{\left(x - 2\right)}^{2} - \frac{7}{2}\right]$

$f \left(x\right) = 2 {\left(x - 2\right)}^{2} - 7$

The vertex is at $\left(2 , - 7\right)$.

Check using the graph:
graph{2x^2-8x+1 [-10, 10, -7.76, 2.24]}