Question

How do you graph `(y-9)^2 = -8(x+5)`?

Answer 1

Please see below.

Explanation:

The equation is of the form `x=a(y-k)^2+h`, whose vertex is `(h,k)` and axis of symmetry is `y-k=0`, as it can be written as

`x=-1/8(y-9)^2-5` ............................(A)

and its vertex is `(-5,9)` and axis of symmetry is `y=9`

Let us select a few points with ordinate around `9`,

say `{1,3,5,7,11,13,15,17}`

Putting these as `y` in (A), we get values of `x` as

`{-13,-9.5,-7,-5.5,-5.5,-7,-9.5,-13}`

Joining these sets of points i.e. `(-13,1)`, `(-9.5,3)`, `(-7,5)`, `(-5.5,7)` `(-5.5,11)`, `(-7,13)`, `(-9.5,15)` and `(-13,17)` and we get the parabola.

graph{(y-9)^2=-8(x+5) [-33.59, 6.41, -1.12, 18.88]}