# Quadratic in vertex form

## 1 Answer

\[f(x)=y= -\frac{1}{2} (x-3)^2 -3 \Rightarrow -\frac{1}{2} (x-3)^2=y+3\]

\[\Rightarrow (x-3)^2=-2(y+3)\]

\[\Rightarrow x-3=\pm\sqrt{-2(y+3)}\]

\[\Rightarrow x=3\pm\sqrt{-2(y+3)}.\]

Note that thif function is not invertable in general, but if we restrict the function $f(x)$ to $x \geq 3$ or $x\leq 3$, then it will be invertable, and the inverse will be obtained by switching the roles of $x$ and $y$ in the above equation:

\[ y=3+\sqrt{-2(x+3)} \text{or} y=3-\sqrt{-2(x+3)},\]

respectively.

Paul F

191

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