In chapter 3, we discussed that every function has an inverse, but only a one-to-one function has an inverse function. Since a quadratic function is not a one-to-one mapping (it is a many-to-one mapping), its inverse is not a function. In other words, the inverse of a many-to-one mapping is a one-to-many mapping, and a one-to-many mapping is not a function.

Graphing the Inverse of a Quadratic Function

We can use a table of values to graph the inverse of a quadratic function by switching the $x$– and $y$-values used in a table of values for the original function. For example, given the function $f(x)=(x+1)^2+5$, we can create a table of values to graph the function (table 1).

The inverse of the function is found by switching the values of the $x$ and $y$ columns so that the inputs are the values of $y$ and the outputs are the values of $x$. Table 2 shows the table of values for the inverse after the values in the $x$– and $y$– columns are switched.

We can use the values in table 1 to graph the original function (in blue) and the values from table 2 to graph the inverse (in green).  Notice that the inverse graph is a reflection of the graph of the original function across the line $y=x$ (in red).

Figure 1. The graph of the inverse of a quadratic function.

Figure 1 shows that the inverse of a quadratic function is also a parabola. However, this parabola is horizontal and opens to the right. Since this horizontal parabola fails the vertical line test, it is not a function.

Restricting the Domain

The graph of a quadratic function is symmetric with a vertical line of symmetry that passes through the vertex of the function. This line of symmetry splits the function into two curves; the curve to the left of the line of symmetry (or the vertex) and the curve to the right. Each of the two curves is one-to-one so the inverse of each curve is a function. Consequently, by restricting the domain to either $x≤h$ or $x≥h$, where $x=h$ is the vertical line of symmetry of the parabola, we create two half-parabolas that are one-to-one and whose inverses are one-to-one functions that are also half-parabolas (figures 2 and 3).

Restricting the domain of a quadratic function $f(x) = (x - h)^2 + k$ to $x≥h$ or $x≤h$ splits the parabola into two parts. The two separate curves are one-to-one and therefore have inverses that are one-to-one functions.

The Inverse Function

To find the inverse function of any quadratic function in the form $f(x) = (x - h)^2 + k$, we must first restrict the domain of the function to make a one-to-one function. We can then switch $x$ and $y$ to get the inverse function. To write this new function using inverse function notation, we solve for $y$ and replace $y$ with $f^{-1}(x)$.

For example, to find the inverse function of $f(x) = (x – 3)^2 + 7$ whose graph has a vertex at (3, 7), we start by restricting the domain to $x \geq 3$ (figure 4). As soon as we switch $x$ and $y$ we have the inverse:

$x=(y-3)^2+7$

Figure 4. Restricting the domain of $f(x)=(x-3)^{2}+7$ to find its inverse function

When we switch $x$ and $y$, the domain of $f(x)$, $\{x\;|\;x≥3\}$, becomes the range of $f^{-1}(x)$ and the range of $f(x)$, $\{f(x)\;|\;f(x)≥7\}$, becomes the domain of $f^{-1}(x)$. This means that the domain of $f^{-1}(x)$ is $\{x\;|\;x≥7\}$ and its range is $\{y\;|\;y≥3\}$ (figure 4).

Now we solve for $y$:

$\begin{aligned}x&=(y-3)^2+7\\ \\x - 7 &= (y - 3)^2\\\\\sqrt{x-7}&=\sqrt{(y-3)^2}\\ \\\sqrt{x-7}&=|y-3|\\\\\sqrt{x-7}&=y-3\;\;\;\;\;\text{Since }y≥3,\;|y-3|=y-3\\\\\sqrt{x-7}+3&=y\end{aligned}$

Therefore, the inverse function of $f(x)=(x-3)^{2}+7$ is $f^{-1}(x) = \sqrt{x - 7} + 3$.

Note that we could have restricted the domain of $f(x) = (x – 3)^2 + 7$ to $\{x\;|\;x≤3\}$ to create a one-to-one function. Figure 5 shows this restricted domain function and its inverse. Compare it to figure 4.

Figure 5. Restricting the domain of $f(x)=(x-3)^{2}+7$ to find its inverse function

Example 2

Find the inverse function of the function $s(x) = (x + 2)^2 – 5$ given $x ≥ –2$.

Solution

We start by writing $y$ for $s(x)$

$s(x) = (x + 2)^2 – 5$

$y = (x + 2)^2 – 5$

Now, exchange the $x$ and $y$.  This also switches the domain and the range, so now $y≥-2$.

$\begin{aligned}x&=(y+2)^2-5\\\\x+5 &= (y + 2)^2\\\\\sqrt{x + 5}&=\sqrt{(y+2)^2}\end{aligned}$

 

Since $y ≥-2$, $y + 2≥0$ and $\sqrt{(y+2)^2}=|y+2|=y+2$:

 

$\begin{aligned}\sqrt{x + 5}&=y+2\\\\\sqrt{x+5}-2&=y\end{aligned}$

 

Therefore, the inverse function of $s(x)=(x + 2)^2 – 5$ is $s^{-1}(x) = \sqrt{x + 5} - 2$.

Try It 2

Find the inverse function of the function $s(x) = (x + 2)^2 – 5$ given $x < –2$.

Show Answer

$s^{-1}(x)=-\sqrt{x + 5} - 2$

Example 3

Find the inverse function of the function $g(x) = 3(x - 4)^2 + 1$ given $x < 4$.

Solution

We start by writing $y$ for $g(x)$ then switching $x$ and $y$. This also switches the domain and the range, so now $y≤4$.

$\begin{aligned}y&= 3(x -4)^2 +1\\\\x&=3(y-4)^2+1\\\\x-1 &= 3(y -4)^2\\\\\dfrac{x-1}{3}&=(y-4)^2\\\\\sqrt{\dfrac{x-1}{3}}&=\sqrt{(y-4)^2}\end{aligned}$

 

Since $y≤4$, $y-4≤0$ and $\sqrt{(y-4)^2}=|y-4|=-(y-4)=-y+4$:

 

$\begin{aligned}\sqrt{\dfrac{x-1}{3}}&=\sqrt{(y-4)^2}\\\\\sqrt{\dfrac{x-1}{3}}&=-y+4\\\\\sqrt{\dfrac{x-1}{3}}-4&=-y\\\\-\sqrt{\dfrac{x-1}{3}}+4&=y\end{aligned}$

 

Therefore, the inverse function of $g(x)=3(x -4)^2 +1$ is $g^{-1}(x) =-\sqrt{\dfrac{x-1}{3}}+4$.

Try It 3

Find the inverse function of the function $g(x) = 3(x - 4)^2 + 1$ given $x ≥ 4$.

Show Answer

$g^{-1}(x) =\sqrt{\dfrac{x-1}{3}}+4$