# How do you find the vertex and intercepts for y = 2(x - 3)^2 + 1?

Apr 1, 2018

vertex is $\left(3 , 1\right)$

Y intercept 19
and
No x intercept

#### Explanation:

In vertex form $f \left(x\right) = A {\left(B \left[x - C\right]\right)}^{2} + D$

We know that C is the x co-ordinate of the vertex and D is the y co-ordinate

So the vertex is $\left(3 , 1\right)$

Y intercept (when x 0)

$y = 2 {\left(\left(0\right) - 3\right)}^{2} + 1 = 2 {\left(- 3\right)}^{2} + 1 = 18 + 1 = 19$

X intercept (when y 0)

$0 = 2 {\left(x - 3\right)}^{2} + 1$
$- 1 = 2 {\left(x - 3\right)}^{2}$
$\sqrt{- 1} = 2 \left(x - 3\right)$

Root 1 doesn't exist on the number line showing that there is no x intercept