# What is the vertex of  y= (x-3)^2-x-2?

Jun 23, 2016

Vertex$\to \left(x , y\right) = \left(\frac{7}{2} , - \frac{45}{2}\right)$

#### Explanation:

Multiply out the bracket so that you combine terms as appropriate.

$y = {x}^{2} - 6 x + 3 \text{ } - x - 2$

$y = {x}^{2} - 7 x + 1$

As the coefficient of ${x}^{2}$ is 1 we can apply directly

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(- 7\right)$ where the -7 is from $- 7 x$

${x}_{\text{vertex}} = + \frac{7}{2}$

Substitute in equation giving

${y}_{\text{vertex}} = {\left(\frac{7}{2}\right)}^{2} - 7 \left(\frac{7}{2}\right) + 1$

${y}_{\text{vertex}} = - 11 \frac{1}{4} \to - \frac{45}{4}$