# How do you find the vertex and intercepts for y=2(x-3)^2+4?

Apr 26, 2017

Vertex is at $\left(3 , 4\right)$. y-intercept is at $\left(0 , 22\right)$ and no x-intercept.

#### Explanation:

Comparing with standard equation y=a(x-h)^2+k ; (h,k) being vertex.
y= 2(x-3)^2+4 ; h=3 , k= 4. So vertex is at $\left(3 , 4\right)$

$y = 2 {\left(x - 3\right)}^{2} + 4 = 2 \left({x}^{2} - 6 x + 9\right) + 4 = 2 {x}^{2} - 12 x + 22$

y-intercept is obtained by putting $x = 0$ in the equation i.e $y = 22$

x-intercept is obtained by putting $y = 0$ in the equation i.e $2 {\left(x - 3\right)}^{2} + 4 = 0 \mathmr{and} 2 {\left(x - 3\right)}^{2} = - 4 \mathmr{and} {\left(x - 3\right)}^{2} = - 2 \mathmr{and} x - 3 = \pm \sqrt{- 2} \therefore x = 3 \pm \sqrt{2} i$
The roots are complex , so no x-intercept is there.

Vertex is at $\left(3 , 4\right)$. y-intercept is at $0 , 22$ and no x-intercept. graph{2x^2-12x+22 [-71.1, 71.1, -35.55, 35.53]}[Ans]