Question

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

Answer 1

see explanation.

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-3)^2+4" is in this form"`

`"with " a=2, h=3" and " k=4`

`rArrcolor(magenta)"vertex "=(3,4)`

`"since " a>0" then min. turning point "uuu`

`color(blue)"Intercepts"`

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

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

`x=0toy=2(-3)^2+4=22larrcolor(red)" y-intercept"`

`y=0to2(x-3)^2+4=0`

`rArr(x-3)^2=-2`

This has no real solutions hence f(x) has no x-intercepts.
graph{2(x-3)^2+4 [-31.56, 31.67, -15.8, 15.8]}