# How do you find the vertex and the intercepts for y = x^2 - 6x + 11?

May 16, 2018

y intercept (0,11)
vertex (3,2)
does not cross the x axis

#### Explanation:

The constant is 11 so the y intercept is (0,11)

complete the square

let ${x}^{2} - 6 x + 11 = 0$

${\left(x - 3\right)}^{2} - 9 + 11 = 0$

${\left(x - 3\right)}^{2} + 2 = 0$

so the x coordinate of the vertex is 3.
Put x=3 into the equation

$y = {3}^{2} - 6 \times 3 + 11$

$y = 9 - 18 + 11$

$y = 2$

So (3,2) is the vertex.

As it is a positive quadratic it is $\cup$ shaped so the vertex is a minimum. As the minimum is above the x axis it does not cross the x axis