# How do you find the vertex and intercepts for y = x^2 – 10x + 5?

Dec 27, 2015

Vertex $\left(5 , - 20\right)$
Intercept $= 5$

#### Explanation:

Vertex

$x = \frac{- b}{2 a} = \frac{- \left(- 10\right)}{2 \times 1} = \frac{10}{2} = 5$

At $x = 5$

$y = {5}^{2} - 10 \left(5\right) + 5 = 25 - 50 + 5 = 30 - 50 = - 20$

Vertex $\left(5 , - 20\right)$
Intercept $= 5$