We start by recalling the key concept: The inverse of a function.

Definition INV Inverse 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