# What is the vertex of  y = (1/4)(4x – 16)^2 - 4?

##### 1 Answer
Apr 15, 2017

Vertex is $\left(4 , - 4\right)$

#### Explanation:

Vertex form of a parabola is $y = a {\left(x + b\right)}^{2} + c$

Notice that the coefficient of $x$ is 1.

In the question asked, the coefficient of $x$ is $4$.

$y = \frac{1}{4} \textcolor{red}{{\left(4 x - 16\right)}^{2}} - 4$

Simplify first: $y = \frac{1}{4} \textcolor{red}{\left(16 {x}^{2} - 128 x + 256\right)} - 4$

Factor out 16:$\text{ }$ (the same as ${4}^{2}$)

$y = \frac{1}{4} \cdot 16 \textcolor{b l u e}{\left({x}^{2} - 8 x + 16\right)} - 4 \text{ } \leftarrow$ change to factor form

$y = 4 \textcolor{b l u e}{{\left(x - 4\right)}^{2}} - 4$

(we could have done this in one step at the beginning as long as the factor ${4}^{2}$ was taken out and not just $4$)

$y = 4 {\left(x - 4\right)}^{2} - 4$ is in vertex form.
The vertex is at $\left(- b , c\right)$

Vertex is $\left(4 , - 4\right)$