We start by recalling the key concept: The inverse of a
function.
Now consider the vertex form of the
quadratic function
$f(x) = A(x-h)^2 + k$.
When $A = 0$, the function $f$ has the same constant value $k$
for any $x$. So $f$ will have no inverse function.
Recall Theorem
QF.SHAPE (The shape of quadratic functions):
It is also apparent from the theorem that when $A > 0$, $f$ is a
one to one from the interval $[h, \infty)$ onto the interval
$[k, \infty)$.
Since all quadratic functions can be constructed using the core
quadratic function, $q(x) =x^2$, and simple linear functions with
composition (Theorem QF.COMP), finding the algebraic expression for the
inverse of $f$ can be interpreted visually by composing the inverses
of $f_{-h}(x)=x-h$, $q(x)=x^2$ , $f_A(x) =A*x$,and
$q_{+k}(x)=x+k$ in the reverse order ("socks and shoes").
Thus the inverse of $f(x) = A(x-h)^2 +k$ is
$g(x)= (f_{+h}\circ q^{-1}\circ f_{1/A}\circ f_{-k})(x) = \sqrt { \frac {x-k} A} +h $.
Therefore $f$ has an "inverse" function,
$g(x) = \sqrt { \frac {x-k} A} +h $, $g
: [k, \infty) \rightarrow [h, \infty)$
and there are comparable results for the interval $( -\infty, h]$
and for when $A < 0$.
This is covered in some beginning and intermediate algebra courses.
In the next examples we use mapping diagrams to check that the
function $g(x) = \sqrt { \frac {x-k} A} +h $ is the "inverse" of the
given function $f$ when $A \gt 0$ by looking at the composition
functions $(f\circ g)(x) = x$ for $x \ge k$ and $(g \circ f)(x) =
x$ for $x \ge h$.
Example
QF.INV.1 :
Suppose $f(x) = 2(x-1)^2 -3$. Verify that $g(x) = \sqrt { \frac
{x+3} 2} + 1$ is the inverse function for $f$ for $x \ge -3$.
"Inverses" for $q_{A*}(x) =A*x^2$
and $q_{+h}(x)=x^2 + h$ are quite simple: $g_{\frac 1 A
*}(x) = \sqrt{ \frac x A }$ and $g_{-h}(x)= \sqrt{x-h}$.
Example
QF.INV.2 :
(i) Suppose $f(x) = 3x^2$. Verify
that $g(x) = \sqrt{ \frac x 3 }$ is the "inverse" function
for $f$.
ii) Suppose $f(x) = x^2 +2$. Verify that $g(x) = \sqrt{x -
2}$ is the "inverse" function for $f$.
Example
QF.INV.3 : Suppose $f(x) = Ax^2 +k$. Verify
that $g(x) =\sqrt{\frac {x-k} A}$ is the inverse function for
$f$.
You can use this next dynamic example
toexamine visually $g$ as the inverse of $f$ with the
mapping diagrams and the graphs of $g$ and $f$.
Example
QF.DINV.0.1 Dynamic Visualization of The Inverse Function
for Quadratic Functions: Graph and Mapping Diagrams