Be careful with finding its inverse! An inverse only exists for one-to-one functions so we need to restrict the domain such that the quadratic is one-to-one. For brevity, we'll take the domain: [tex]x \geq 0[/tex] but you could just as easily take [tex]x \leq 0[/tex].
Being that the quadratic is now one-to-one, we'll try to find its inverse.
By definition, an inverse function is the function flipped across the line y = x. So being that, we need to interchange the x and y coordinates around and then solve for y.
This gives us:
[tex]x = -4y^2 - 2 \\ x + 2 = -4y^2 \\ -\frac{1}{4}\left(x + 2\right) = y^2 \\ y = \sqrt{-\frac{1}{4}\left(x + 2\right)}[/tex]
Now, this function is defined for when [tex]y \geq 0[/tex] and [tex]x \leq -2[/tex].
So I'll end with this, what is the relationship between the domain/range of the original function and the domain/range of the inverse?
