# What is the vertex form of y= x^2/4 - x - 4 ?

Aug 27, 2017

$y = \frac{1}{4} {\left(x - 2\right)}^{2} - 5$

#### Explanation:

The given equation

$y = {x}^{2} / 4 - x - 4 \text{ [1]}$

is in standard form:

$y = a {x}^{2} + b x + c$

where $a = \frac{1}{4} , b = - 1 \mathmr{and} c = - 4$

Here is a graph of the given equation:

graph{x^2/4 - x - 4 [-8.55, 11.45, -6.72, 3.28]}

The vertex form for a parabola of this type is:

$y = a {\left(x - h\right)}^{2} + k \text{ [2]}$

where $\left(h , k\right)$ is the vertex.

We know that "a" in the standard form is the same as the vertex form, therefore, we substitute $\frac{1}{4}$ for "a" into equation [2]:

$y = \frac{1}{4} {\left(x - h\right)}^{2} + k \text{ [3]}$

To find the value of $h$, we use the formula:

$h = - \frac{b}{2 a}$

Substituting in the values for "a" and "b":

$h = - \frac{- 1}{2 \left(\frac{1}{4}\right)}$

$h = 2$

Substitute 2 for $h$ into equation [3]:

$y = \frac{1}{4} {\left(x - 2\right)}^{2} + k \text{ [4]}$

To find the value of k, we evaluate the given equation at $x = h = 2$:

$k = {\left(2\right)}^{2} / 4 - 2 - 4$

$k = 1 - 2 - 4$

$k = - 5$

Substitute -5 for $k$ into equation [4]:

$y = \frac{1}{4} {\left(x - 2\right)}^{2} - 5$

Here is a graph of the vertex form:

graph{1/4(x-2)^2-5 [-8.55, 11.45, -6.72, 3.28]}

Please observe that the two graphs are identical.