# How do you rewrite into vertex form f(x)=4x^2-16x+5?

Nov 14, 2016

#### Explanation:

The vertex form for a parabola that opens or down is:

f(x) = a(x - h)^2 + k

Please observe that, in the given equation, $a = 4$, therefore we add zero to the original equation in the form of $4 {h}^{2} - 4 {h}^{2}$:

$f \left(x\right) = 4 {x}^{2} - 16 x + 4 {h}^{2} - 4 {h}^{2} + 5$

Factor 4 out of the first 3 terms and group them with ()s:

$f \left(x\right) = 4 \left({x}^{2} - 4 x + {h}^{2}\right) - 4 {h}^{2} + 5$

We can find the value of h by setting the middle term equal to -2hx:

$- 2 h x = - 4 x$

$h = 2$ and $- 4 {h}^{2} = - 16$

$f \left(x\right) = 4 \left({x}^{2} - 4 x + {h}^{2}\right) - 16 + 5$

We know that three terms inside the ()s is equal to ${\left(x - 2\right)}^{2}$

$f \left(x\right) = 4 {\left(x - 2\right)}^{2} - 11$