How do you find the vertex for 2x^-16x+4?

Jun 27, 2015

I think you mean $2 {x}^{2} - 16 x + 4$

$2 {x}^{2} - 16 x + 4 = 2 {\left(x - 4\right)}^{2} - 28$

This is in vertex form allowing us to read the vertex as $\left(4 , - 28\right)$

Explanation:

Let $f \left(x\right) = 2 {x}^{2} - 16 x + 4$

Then $f \left(x\right) = 2 \left({x}^{2} - 8 x + 16\right) - 32 + 4 = 2 {\left(x - 4\right)}^{2} - 28$

$f \left(x\right) = 2 {\left(x - 4\right)}^{2} - 28$ is in vertex form, being of the form:

$f \left(x\right) = a \left(x - {x}_{1}\right) + {y}_{1}$,

where $\left({x}_{1} , {y}_{1}\right) = \left(4 , - 28\right)$ is the vertex and $a$ is some constant.

More explicitly:

The minimum value of $f \left(x\right)$ occurs when $\left(x - 4\right) = 0$, that is when $x = 4$.

$f \left(4\right) = 2 {\left(4 - 4\right)}^{2} - 28 = 0 - 28 = - 28$

So the vertex is at $\left(4 , - 28\right)$