# How do you find the inverse of f(x)=x^2-6x?

Oct 14, 2015

${f}^{-} 1 \left(x\right) = \sqrt{x + 9} + 3$

#### Explanation:

First you equate $f \left(x\right)$ to another variable, say $f \left(x\right) = y$.

But before that, complete the square for $f \left(x\right)$.

$\therefore {x}^{2} - 6 x + {3}^{2} - {3}^{2} = {\left(x - 3\right)}^{2} - 9$

Now,
$y = {\left(x - 3\right)}^{2} - 9$
${\left(x - 3\right)}^{2} = y + 9$
$\left(x - 3\right) = \sqrt{y + 9}$
$x = \sqrt{y + 9} + 3$

Now swap back the x for the variable y.

So the inverse for $f \left(x\right)$ is
${f}^{-} 1 \left(x\right) = \sqrt{x + 9} + 3$