Question

How do you find the vertex and intercepts for `y = 2(x – 5)^2 + 2`?

Answer 1

`(5,2),(0,52)" no x-intercepts"`

Explanation:

The equation of a parabola in `color(blue)"vertex form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))`
where (h , k) are the coordinates of the vertex and a is a constant.

`y=2(x-5)^2+2" is in this form"`

`"with " (h,k)=(5,2)larrcolor(red)" vertex"`

`color(blue)" to find intercepts"`

`• " let x=0, in equation, for y-intercept"`

`• " let y=0, in equation, for x-intercept"`

`x=0toy=2(-5)^2+2=52larrcolor(red)" y-intercept"`

`y=0to2(x-5)^2+2=0`

`rArr(x-5)^2=(-2)/2=-1`

`(x-5)^2=-1" has no real solutions thus no x-intercepts"`
graph{2(x-5)^2+2 [-10, 10, -5, 5]}