# How do you find the vertex for y = x^2 - 2x?

Apr 3, 2018

The vertex is at $\left(1 , - 1\right)$

#### Explanation:

We can quite easily see where the vertex of the quadratic function is if we write it in vertex form:

$a {\left(x - h\right)}^{2} + k$ with vertex at $\left(h , k\right)$

To complete the square, we need $h$ to be half the $x$ coefficient, so in this case we have $- 2 \frac{\setminus}{2} = - 1$:

${\left(x - 1\right)}^{2} + k = {x}^{2} - 2 x$

${x}^{2} - 2 x + 1 + k = {x}^{2} - 2 x$

$k = - 1$

This means the vertex form of our quadratic function is:

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

And therefore the vertex is at $\left(1 , - 1\right)$