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

May 17, 2018

Vertex form is $y = {\left(x - 1\right)}^{2} - 6$

The vertex is the point $\left(1 , - 6\right)$

#### Explanation:

Vertex form is $y = a {\left(x + p\right)}^{2} + q$ with the vertex at $\left(- p , q\right)$

This is derived by the process of completing the square.

The quadratic trinomial ${x}^{2} - 2 x - 5$ is in the form $a {x}^{2} + b x + c$

${x}^{2} - 2 x - 5$ is not a perfect square.

We need to add the correct constant which will make it a square.
This is found from ${\left(\frac{b}{2}\right)}^{2}$ which in this case is ${\left(\frac{- 2}{2}\right)}^{2} = \textcolor{b l u e}{1}$

$y = {x}^{2} - 2 x \textcolor{b l u e}{+ 1} \textcolor{b l u e}{- 1} - 5 \text{ } \leftarrow \left(\textcolor{b l u e}{+ 1 - 1 = 0}\right)$

$y = \left({x}^{2} - 2 x \textcolor{b l u e}{+ 1}\right) + \left(\textcolor{b l u e}{- 1} - 5\right)$

$y = {\left(x - 1\right)}^{2} - 6 \text{ } \leftarrow$ this is vertex form.

The vertex is the point $\left(1 , - 6\right)$