Question

How do you graph `f(x)= -(x+4)^2 + 9`?

Answer 1

see explanation.

Explanation:

The following points are necessary.

`• " coordinates of vertex"`

`• " x and y intercepts"`

`• " shape of parabola. That is max / min"`

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.

`f(x)=-(x+4)^2+9" is in this form"`

`"with " a=-1, h=-4" and " k=9`

`rArrcolor(magenta)"vertex "=(-4,9)`

`"since " a<0" then max turning point " nnn`

`color(blue)"intercepts"`

`x=0toy=-16+9=-7larrcolor(red)" y-intercept"`

`y=0to-(x+4)^2+9=0`

`rArr(x+4)^2=9`

`rArrx+4=+-3`

`rArrx=-1" or " x=-7larrcolor(red)"x-intercepts"`
graph{-(x+4)^2+9 [-20, 20, -9.98, 10.02]}