# What is the vertex form of y= x^2 – 5x – 6 ?

Apr 29, 2017

$y = {\left(x - \frac{5}{2}\right)}^{2} - \frac{49}{4}$

#### Explanation:

The equation of a parabola in $\textcolor{b l u e}{\text{vertex form}}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where (h ,k) are the coordinates of the vertex and a is a constant.

$\text{using the method of "color(blue)"completing the square}$

add (1/2" coefficient of x-term")^2" to " x^2-5x

Since we are adding a value which is not there we must also subtract this value.

$\text{add / subtract } {\left(- \frac{5}{2}\right)}^{2} = \frac{25}{4}$

$y = \left({x}^{2} - 5 x \textcolor{red}{+ \frac{25}{4}}\right) \textcolor{red}{- \frac{25}{4}} - 6$

$\textcolor{w h i t e}{y} = {\left(x - \frac{5}{2}\right)}^{2} - \frac{49}{4} \leftarrow \textcolor{red}{\text{ in vertex form}}$