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):**

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, 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 $x \ge h$.

"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 $f$.

ii) Suppose $f(x) = x^2 +2$. Verify that $g(x) = \sqrt{x - 2}$ 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 $f$.