Quadratic in vertex form
can anyone help me find the inverse of the function f(x)= -1/2 (x-3)^2 -3 ? This is a quad in vertex form if confused.
1 Answer
We have
\[f(x)=y= -\frac{1}{2} (x-3)^2 -3 \Rightarrow -\frac{1}{2} (x-3)^2=y+3\]
\[\Rightarrow (x-3)^2=-2(y+3)\]
\[\Rightarrow x-3=\pm\sqrt{-2(y+3)}\]
\[\Rightarrow x=3\pm\sqrt{-2(y+3)}.\]
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.
\[f(x)=y= -\frac{1}{2} (x-3)^2 -3 \Rightarrow -\frac{1}{2} (x-3)^2=y+3\]
\[\Rightarrow (x-3)^2=-2(y+3)\]
\[\Rightarrow x-3=\pm\sqrt{-2(y+3)}\]
\[\Rightarrow x=3\pm\sqrt{-2(y+3)}.\]
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
200
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