Question

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

Answer 1

Vertex is at `(3,4)`. y-intercept is at `(0,22)` 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 `(3,4)`

`y= 2(x-3)^2+4 = 2(x^2-6x+9)+4 = 2x^2-12x+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(x-3)^2+4=0 or 2(x-3)^2 =-4 or (x-3)^2= -2 or x-3 = +-sqrt(-2) :. x=3+-sqrt 2i`
The roots are complex , so no x-intercept is there.

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