# How do you graph y = (x - 1)^2 + 2?

May 20, 2018

$\left(1 , 2\right)$ is the vertex

#### Explanation:

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

This function is in vertex form:

$y = a \left(x - h\right) + k$

$\left(- h , k\right) = \left(1 , 2\right)$ is the vertex

to find the y-intcept set x=0 and solve:

$y = {\left(0 - 1\right)}^{2} + 2$

$y = 3$

to find the x-intcepts set y=0 and solve:

$0 = {\left(x - 1\right)}^{2} + 2$

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

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

$x - 1 = \pm \sqrt{- 2}$

$x = 1 \pm \sqrt{2} i$. so there are no real roots hence no x-intercepts.

graph{(x - 1)^2 + 2 [-17.48, 23.07, -0.65, 19.63]}