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 [latex]x[/latex]– and [latex]y[/latex]-values used in a table of values for the original function. For example, given the function [latex]f(x)=(x+1)^2+5[/latex], 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 [latex]x[/latex] and [latex]y[/latex] columns so that the inputs are the values of [latex]y[/latex] and the outputs are the values of [latex]x[/latex]. Table 2 shows the table of values for the inverse after the values in the [latex]x[/latex]– and [latex]y[/latex]– 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 [latex]y=x[/latex] (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 [latex]x≤h[/latex] or [latex]x≥h[/latex], where [latex]x=h[/latex] 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).

Figure 2. Parabola and its inverse with restricted domain [latex]x≥h[/latex]	Figure 3. Parabola and its inverse with restricted domain [latex]x≤h[/latex]

Restricting the domain of a quadratic function [latex]f(x) = (x - h)^2 + k[/latex] to [latex]x≥h[/latex] or [latex]x≤h[/latex] 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 [latex]f(x) = (x - h)^2 + k[/latex], we must first restrict the domain of the function to make a one-to-one function. We can then switch [latex]x[/latex] and [latex]y[/latex] to get the inverse function. To write this new function using inverse function notation, we solve for [latex]y[/latex] and replace [latex]y[/latex] with [latex]f^{-1}(x)[/latex].

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

[latex]x=(y-3)^2+7[/latex]

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

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

Now we solve for [latex]y[/latex]:

[latex]\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}[/latex]

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

Note that we could have restricted the domain of [latex]f(x) = (x – 3)^2 + 7[/latex] to [latex]\{x\;|\;x≤3\}[/latex] 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 [latex]f(x)=(x-3)^{2}+7[/latex] to find its inverse function

Example 2

Find the inverse function of the function [latex]s(x) = (x + 2)^2 – 5[/latex] given [latex]x ≥ –2[/latex].

Solution

We start by writing [latex]y[/latex] for [latex]s(x)[/latex]

[latex]s(x) = (x + 2)^2 – 5[/latex]

[latex]y = (x + 2)^2 – 5[/latex]

Now, exchange the [latex]x[/latex] and [latex]y[/latex].  This also switches the domain and the range, so now [latex]y≥-2[/latex].

[latex]\begin{aligned}x&=(y+2)^2-5\\\\x+5 &= (y + 2)^2\\\\\sqrt{x + 5}&=\sqrt{(y+2)^2}\end{aligned}[/latex]

 

Since [latex]y ≥-2[/latex], [latex]y + 2≥0[/latex] and [latex]\sqrt{(y+2)^2}=|y+2|=y+2[/latex]:

 

[latex]\begin{aligned}\sqrt{x + 5}&=y+2\\\\\sqrt{x+5}-2&=y\end{aligned}[/latex]

 

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

Try It 2

Find the inverse function of the function [latex]s(x) = (x + 2)^2 – 5[/latex] given [latex]x < –2[/latex].

Show Answer

[latex]s^{-1}(x)=-\sqrt{x + 5} - 2[/latex]

Example 3

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

Solution

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

[latex]\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}[/latex]

 

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

 

[latex]\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}[/latex]

 

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

Try It 3

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

Show Answer

[latex]g^{-1}(x) =\sqrt{\dfrac{x-1}{3}}+4[/latex]