Question

How do you find the vertex and intercepts for `y = 2(x + 4)^2 - 21`?

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+4)^2-21" is in this form"`

`"with " h=-4" and " k=-21`

`rArrcolor(magenta)"vertex "=(-4,-21)`

`color(blue)"to find intercepts"`

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

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

`x=0toy=2(4)^2-21=32-21=11larrcolor(red)" y-intercept"`

`y=0to2(x+4)^2-21=0`

`rArr2(x+4)^2=21`

`rArr(x+4)^2=21/2`

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

`sqrt((x+4)^2)=color(red)(+-)sqrt(21/2)larr" note plus or minus"`

`rArrx+4=+-sqrt(21/2)`

`"subtract 4 from both sides"`

`xcancel(+4)cancel(-4)=+sqrt(21/2)-4`

`rArrx=-4+-sqrt(21/2)larrcolor(red)" x-intercepts"`

`"these are the exact values, which may be approximated to"`

`x=-7.24" and " x=-0.76" to 2 dec. places"`
graph{2(x+4)^2-21 [-46.3, 46.16, -23.07, 23.18]}