Question

How do you find the vertex and intercepts for `y=2(x+2)^2-9`?

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+2)^2-9" is in this form"`

`"with " h=-2" and " k=-9`

`rArrcolor(magenta)"vertex "=(-2,-9)`

`color(blue)"to find intercepts"`

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

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

`x=0toy=2(2)^2-9=-1larrcolor(red)" y-intercept"`

`y=0to2(x+2)^2-9=0`

`rArr(x+2)^2=9/2`

`color(blue)"taking the square root of both sides"`

`sqrt((x+2)^2)=color(red)(+-)sqrt(9/2)larr" note plus or minus"`

`rArrx+2=+-3/sqrt2`

`rArrx=-2+-(3sqrt2)/2`

`x=-2+(3sqrt2)/2~~0.12larrcolor(red)" x-intercept"`

`x=-2-(3sqrt2)/2~~-4.12larrcolor(red)" x-intercept"`
graph{2(x+2)^2-9 [-28.87, 28.86, -14.43, 14.44]}