# How do you find the vertex and the intercepts for y = x^2 – 4x + 9?

May 8, 2018

vertex is (2,5)
(0,9) is the y intercept
no x intercepts

#### Explanation:

Complete the square

$y = {\left(x - 2\right)}^{2} - 4 + 9$

$y = {\left(x - 2\right)}^{2} + 5$

so the vertex is (2,5)

If $y = {x}^{2} - 4 x + 9$ when $x = 0 , y = 9$

If we take the equation after we have completed the square

$y = {\left(x - 2\right)}^{2} + 5$ and put $y = 0$

$0 = {\left(x - 2\right)}^{2} + 5$

${\left(x - 2\right)}^{2} = - 5$

$x - 2 = \sqrt{- 5}$

As we cannot take the square root of a negative number there are no roots so it does not cross the $y$ axis