Find the inverse $f(x) = 2x^2-8x, x>2 $

Question

$$ 2x^2-8x, x>2 $$

What is the best way to solve this problem.

$$x = 2y^2-8y $$

$$x = y (2y-8) $$ do I divide both sides by $y$ so as to solve for $y$? Help

Answer 1

$y=2x^2-8x$

so

$\frac{y}{2}=x^2-4x=(x-2)^2-4$

so (since $x>2$)

$x=2+\sqrt{4+\frac{y}{2}}$