Alternative Format — Lesson 3: Inverses of Quadratic Functions

Let's Start Thinking

Inverse Functions

For example, consider \(f(x)=3x+1\):

We can visualize this function as a machine.

Then the inverse of \(f\) must be a relation that undoes this work.

Previously, we have focused mainly on linear functions.

In this lesson, we will spend most of our time looking at quadratic functions and their inverses.

Try This

The inverse of \(f(x) = -2(x-4)^2+7\) is not a function. However, it is possible to restrict the domain of \(f(x)\) so that its inverse is a function.

Determine a way to restrict the domain of \(f(x)\) so that the inverse of \(f(x)\) is a function.

When Is an Inverse Relation a Function?

The Vertical Line Test Revisited

The Vertical Line Test is used to determine if a relation, represented graphically, is a function. If there exists a value of \(x\) in the domain of the relation for which there are at least two values of \(y\), then the relation is not a function. Equivalently, the Vertical Line Test states:

Explore This 1

Question — Version 1

Is the inverse of \(f(x)\) also a function?

Use the vertical line test to help determine if the inverse of \(f(x)\) is a function.

Answer — Version 1

The inverse is not a function.

Question — Version 2

Is the inverse of \(f(x)\) also a function?

Use the vertical line test to help determine if the inverse of \(f(x)\) is a function.

Answer — Version 2

The inverse is not a function.

Question — Version 3

Is the inverse of \(f(x)\) also a function?

Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0.

Use the vertical line test to help determine if the inverse of \(f(x)\) is a function.

Answer — Version 3

The inverse is not a function.

Interactive Version

Vertical Line Test

Explore This 1 Summary

Perhaps you noticed the following in Explore This 1:

• When \(f(x)\) is a non-horizontal linear function, the inverse of \(f(x)\) is a function.
• When \(f(x)\) is a horizontal linear function, the inverse of \(f(x)\) is not a function.
• When \(f(x)\) is a quadratic function, the inverse of \(f(x)\) is not a function.

One-to-One and Many-to-One Functions

If we think in terms of inputs into the function and outputs of the function:

• A function is one-to-one if each output corresponds to exactly one input. Since we are describing a function, it is also true that each input corresponds to exactly one output.
• A function is many-to-one if there is at least one output that corresponds to more than one input. For instance, the function \(f(x)=x^2\) is many-to-one since we can find an output value, say \(f(x)=4\), that corresponds to more than one input (since \(f(2)=4\) and \(f(-2)=4\)).

Let's see how this works on a mapping diagram.

Notice that \(f(x)\) and \(g(x)\) are both functions: each input has exactly one output.

The function \(f(x)\) is one-to-one since each value of the dependent variable corresponds to exactly one value of the independent variable.

The function \(g(x)\)  is many-to-one since there are two input values, \(x=2\) and \(x=-2\), that map to a single output value, \(4\). Thus, there is a value of the dependent variable that corresponds to different values of the independent variable.

Now consider the inverses of \(f(x)\) and \(g(x)\):

Notice that the inverse of \(f(x)\) is a function since each input has exactly one output. However, the inverse of \(g(x)\) is not a function since there is now an input value, \(x=4\), that maps to two output values, \(2\) and \(-2\).

In general, we can state:

Let's continue looking at when the inverse of a function is also a function.

Example 1

Solution — Graph A

To answer part a), let's start by graphing the inverse of \(f(x)\).

To graph the inverse, we will interchange the \(x\) and \(y\)-coordinates for several key points along \(y = f(x)\).

• \((-4,0)\) maps to \((0,-4)\)

  

• \((0, 2)\) maps to \((2, 0)\)

  

As we have seen previously, since these points, \((-4, 0)\) and \((0, 2)\), on \(f(x)\) are connected by a straight line, the image points, \((0, -4)\) and \((2, 0)\), on the inverse will be as well.

Continuing this process, we get the graph of the inverse as shown.

So is the inverse a function? Let's apply the vertical line test.

If there is a vertical line that crosses the inverse at more than one point, then the inverse is not a function. Indeed consider this line drawn at \(x = 1\).

It crosses the inverse at three points. So the inverse fails the vertical line test. There is an input that has more than one output.

Thus, the inverse in this case is not a function and this is why the notation of \(f^{-1}(x)\) has not been used in this question.

Here's an observation that will speed things up significantly for us in these types of questions.

We concluded that the inverse was not a function because the graph of the inverse failed the vertical line test. Here we see again that a vertical line drawn at \(x = 1\) crosses the inverse in three places.

Let's track that back to the original function.

Since \(x\) and \(y\) interchanged roles,

Checking if a vertical line intersects the inverse graph in more than one point

is equivalent to

Checking if a horizontal line intersects the original function at more than one point.

Take a look at the original function again and consider the horizontal line \(y = 1\).

Like its vertical line counterpart used on the inverse function, this horizontal line intersects the original function at three points.

Solution — Graph B

Let's use this idea to complete graph b).

Is the inverse of this function also a function?

Notice that any horizontal line we draw will intersect the original function at only one point.

Consequently, the inverse is a function.

We can confirm this by looking at the graph of the inverse and noticing that there are no vertical lines that would intersect the graph in more than one point.

Check Your Understanding 1

Question — Version 1

Given the graph of \(f(x)\) determine if the inverse of \(f(x)\) is a function.

Seeing the inverse

Answer — Version 1

The inverse of \(f\) is a function.

Feedback — Version 1

The inverse of \(f\) is a function. A vertical line cannot be drawn in such a way that it intersects the graph of the inverse in more than one point. 

Equivalently, a horizontal line cannot be drawn in such a way that it intersects the graph of \(f\) in more than one point.

Question — Version 2

Given the graph of \(f(x)\) determine if the inverse of \(f(x)\) is a function.

Seeing the inverse

Answer — Version 2

The inverse of \(f\) is a function.

Feedback — Version 2

The inverse of \(f\) is a function. A vertical line cannot be drawn in such a way that it intersects the graph of the inverse in more than one point. 

Equivalently, a horizontal line cannot be drawn in such a way that it intersects the graph of \(f\) in more than one point.

Question — Version 3

Given the graph of \(f(x)\) determine if the inverse of \(f(x)\) is a function.

Seeing the inverse

Answer — Version 3

The inverse of \(f\) is not a function.

Feedback — Version 3

Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0.

The inverse of \(f\) is not a function. A vertical line can be drawn that intersects the graph of the inverse in more than one point. 

Equivalently, a horizontal line can be drawn that intersects the graph of \(f\) in more than one point.

Interactive Version

Functions and their Inverses

Determining the Inverse of a Quadratic Function Algebraically

Before seeing an example of calculating the inverse of a quadratic function algebraically, recall the following steps used previously to calculate the inverse of a linear function algebraically.

Steps to Determine an Inverse Algebraically

When \(f(x)\) is defined algebraically, the steps to determine its inverse are:

1. Rewrite \(f(x)\) as \(y\).
2. Write "To determine the inverse, interchange \(x\) and \(y\):"
3. Interchange \(x \) and \(y\).
4. Rearrange* the equation to isolate for \(y\).
5. If the inverse is a function, rewrite \(y\) as \(f^{-1}(x)\).

*In this unit, mainly quadratic and non-horizontal linear functions \(f(x)\) are considered; rearrangement is always possible with such functions.

Let's try an example of algebraically finding the inverse of a quadratic relation.

Example 2

Solution

Here are the steps we will use to determine the inverse of \(f(x)\).

\( f(x) =-2(x-4)^{2}+7 \)

\(y =-2(x-4)^{2}+7 \)

1. Rewrite \(f(x) \) as \(y\).
2. Write "To determine the inverse, interchange \(x\) and \(y\):".
3. Interchange \(x\) and \(y\).

   \(x = -2(y-4)^2+7\)

   Remember that at this stage the meaning of \(y\) has changed. Before this point, \(y\) represents \(f(x)\). From this point forward, \(y\) represents \(f\) inverse of \(x\).  That's why we include this phrase in the solution, interchange \(x\) and \(y\), to show that we have changed the function.

4. Isolate for \(y\) using inverse operations.

   We subtract \(7\) from both sides.

   \(x-7 =-2(y-4)^{2}\)

   Then divide both sides by \(-2\).

   \((y-4)^{2} =\dfrac{x-7}{-2} \)

   We've also switched the left side and right side of the equation at this point, just to get \(y\) on the left side.

   Then to get rid of the negative in the denominator, let's multiply numerator and denominator by \(-1\).

   \((y-4)^{2} =\dfrac{-x+7}{2}\)

   Then we'll take the square root of both sides, remembering to include both the positive and negative square root.

   \( y-4 =\pm \sqrt{\dfrac{7-x}{2}} \)

   Also, \(-x + 7\) has been rewritten in opposite order as \(7 - x\).

    

   Finally, we add \(4\) to both sides.

   \(y =4 \pm \sqrt{\dfrac{7-x}{2}} \)

5. In the last step, if the inverse is a function, rewrite \(y\) as \(f^{-1}(x)\).

   So is \(y =4 \pm \sqrt{\dfrac{7-x}{2}} \) a function?

   No, it is not. Because of the \(\pm\), there are inputs that have multiple outputs.

   E.g., when \(x = -1\), we can evaluate \(y\) as shown.

   \(\begin{align*} y &=4 \pm \sqrt{\dfrac{7+1}{2}} \\ &=4 \pm \sqrt{4} \\ &=4 \pm 2 \\ &=2 \text { or } 6 \end{align*}\)

   Therefore, the inverse of \(f\) is not a function.

   So we will not use function notation for the inverse.

Let's look at a graph of \( f(x) =-2(x-4)^{2}+7 \) and its inverse.

Since \(f(x)\) is written in vertex form, we know that \(f(x)\) has vertex \((4, 7)\), it opens down, and it has been stretched vertically by a factor of \(2\).

So here is a graph of \(y = f(x)\).

In this particular example, we could also graph the inverse directly, using the equation of the inverse we found previously. This would require that we would graph two square root functions.

We'll proceed now by reflecting \(y = f(x)\) in the line \(y = x\).

The vertex of \(y = f(x)\), \((4, 7)\), will map to \((7, 4)\) and this will be the vertex of our reflected parabola.

We could use the equation to see that two points on \(y = f(x)\) are \((2, -1)\), and \((6, -1)\).

These map to \((-1, 2)\), and \((-1, 6)\) on the inverse.

Here are two other points on the original function and their images on the inverse.

So we begin to see that the inverse is a parabola opening to the left.

Notice that the inverse does not pass the vertical line test.

That's expected, since we already saw that the inverse is not a function.

Check Your Understanding 2

Question

Determine the inverse of \(f(x)=-2(x+3)^2 +3 \).

Answer

\(y = -3 \pm \sqrt{\dfrac{x-3}{-2}}\)

Feedback

\(\begin{align*} f(x) &=-2(x+3)^2 +3 \\ y&=-2(x+3)^2 +3 \end{align*}\)

To determine the inverse, interchange \(x\) and \(y\):

\(\begin{align*} x &=-2(y+3)^2 +3 \\ x-3 &=-2(y+3)^2 \\ -2(y+3)^2 &= x-3 \\ (y+3)^2 &= \dfrac{x-3}{-2} \\ y+3 &= \pm \sqrt{\dfrac{x-3}{-2}} \\ y &= -3 \pm \sqrt{\dfrac{x-3}{-2}} \end{align*}\)

Example 3

Solution — Part A

We will need to interchange \(x\) and \(y\) to determine an equation for the inverse. If we interchange variables in the equation \(y=3(x-4)(x+2)\), we arrive at \(x=3(y-4)(y+2)\).

Solving this equation for \(y\) is very challenging; a better approach is to write \(f(x)\) in vertex form before interchanging \(x\) and \(y\).

We proceed using these two stages:

• Stage 1: Convert to vertex form
• Stage 2: Determine the inverse

Stage 1: Convert to vertex form

Since \(f(x)=3(x-4)(x+2)\) is written in factored form, we observe that

• The zeros are \(4\) and \(-2\).
• The axis of symmetry, which lies halfway between the zeros, is \(x=1\).
• The \(y\)-coordinate of the vertex is \(f(1)\) where

  \(\begin{align*} f(x)&=3(x-4)(x+2)\\ f(1)&=3(1-4)(1+2)\\ &= -27\\ \end{align*}\)

• The vertex is \((1,-27)\).

Therefore, the vertex form equation is \(f(x)=3(x-1)^2-27\) (since \(a=3\) is the same in both factored form and vertex form).

Stage 2: Determine the inverse

\(\begin{align*} f(x)&=3(x-1)^2-27\\ y&=3(x-1)^2-27\\ \end{align*}\)

To determine the inverse, interchange \(x\) and \(y\):

\(\begin{align*} x&=3(y-1)^2-27\\ x+27&=3(y-1)^2\\ (y-1)^2&=\dfrac{x+27}{3}\\ y-1&=\pm\sqrt{\dfrac{x+27}{3}}\\ y&=1\pm\sqrt{\dfrac{x+27}{3}} \end{align*}\)

Therefore, the inverse of \(f(x)=3(x-4)(x+2)\) is given by \(y = 1\pm\sqrt{\dfrac{x+27}{3}}\).

Solution — Part B

Determine the inverse of \(g(x)=3x^2-6x-24\).

We proceed using these two stages:

• Stage 1: Convert to vertex form
• Stage 2: Determine the inverse

Stage 1: Convert to vertex form

Since \(g(x)\) is written in standard form, we will convert it to vertex form by completing the square:

Stage 2: Determine the inverse

Notice that \(g(x)\) is the same function as in part (a). Therefore, the inverse of \(g(x)=3x^2-6x-24\) is given by \(y = 1\pm\sqrt{\dfrac{x+27}{3}}\).

Restricting the Domain of a Quadratic Function

Terminology

We similarly can speak of the upper arm and lower arm of a parabola opening to the left or to the right.

As we move on to restricting the domain of a quadratic function so that the inverse is a function, let's revisit the Try This problem presented at the beginning of this lesson.

Try This Revisited

Solution

In a previous example, we found this graph of \(y = f(x)\) and its inverse.

The inverse of \(f(x)\) is not a function since it fails the Vertical Line Test.

For example, the points \((-1,2)\) and \((-1,6)\) both satisfy the inverse relation.

We found earlier that an equation for the inverse is \(y=4\pm\sqrt{\dfrac{7-x}{2}}\).

We also see from the equation that this relation is not a function. The \(\pm\) gives two different outputs for many inputs.

This relation is the union of two separate functions

\(\class{hl4}{y=4 + \sqrt{\dfrac{7-x}{2}}}\)

\(\class{hl2}{y=4 - \sqrt{\dfrac{7-x}{2}}}\)

The equation with the plus sign corresponds to the upper arm of the parabola, and the equation with the minus sign generates the lower arm of the parabola.

You can check this by substituting \(x = -1\) into each equation to see which equation generates \(2\), and which equation generates \(6\).

Now remember the goal of this try this.

We're trying to find a restricted domain of the original function, so that the inverse function is indeed a function.

So if we were to only have this part of the inverse, then the inverse would be a function. So let's reverse engineer that to see which section of \(y = f(x)\) corresponds to this part of the inverse.

If we reflect this upper arm in the line \(y = x\), then we arrive at the right arm of \(y = f(x)\).

Therefore, if we restrict the domain of \(f(x)\) to include only the right arm, then the inverse consists only of the upper arm, so that it is a function.

The domain that generates the right arm of \(y = f(x)\) is \(D=\{x\in \mathbb{R} \mid x\ge 4 \}\).

Alternatively, we could also plan to restricted domain on \(f(x)\), so that only the lower arm of the inverse relation parabola remains.

This corresponds to the domain that retains only the left arm of \(f(x)\).

In that case, \(D= \{x\in \mathbb{R} \mid x\le 4 \}\).

Thus, we have two possible restricted domains of \(f(x)\), that ensure that the inverse is a function.

This restricted domain corresponds to the right arm of \(f(x)\) in the upper arm of the inverse. So if \(f(x)\) is restricted or limited to this domain, then its inverse is a function.

Notice that, as we have seen previously the domain of \(f(x)\), is the range of the inverse of \(f(x)\).

Likewise, this domain corresponds to the left arm of \(f(x)\) and the lower arm of the inverse.

Again, the domain of \(f(x)\) is the same as the range of the inverse.

There are other restricted domains that ensure that the inverse is a function but these are the only two that retain one complete arm of the parabola.

By ensuring that one complete arm of the inverse is retained, we keep as many key features of the parabola as possible, while also ensuring that the inverse is a function.

This will always be the intention in these questions.

Restricting the Domain of a Quadratic Function

Suppose a quadratic function \(f(x)=ax^2+bx+c\) has vertex \((h,k)\). Then, the inverse of \(f(x)\) is not a function.

However, if the domain of \(f(x)\) is restricted to either \(D=\{x\in\mathbb{R} \mid x \ge h\}\) or \(D=\{x\in\mathbb{R} \mid x \le h\}\) then, the inverse

• is a function, and
• will contain one full arm of the parabola.

Example 4

Solution — Part A

Since \(f(x)=-\dfrac15(x+3)(x-5)\) is in factored form, we observe that:

• The zeros are \(-3\) and \(5\).
• The axis of symmetry, which lies halfway between the zeros, is \(x=1\).
• The \(x\)-coordinate of the vertex is \(1\) since the vertex lies on the axis of symmetry.

Therefore, either \(D=\{x\in\mathbb{R} \mid x\ge1\}\) or \(D=\{x\in\mathbb{R} \mid x\le1\}\) is a restricted domain that ensures the inverse is a function and contains one complete arm of the parabola.

Solution — Part B

Since \(g(x)=x^2+6x+4\) is in standard form, we first convert it to vertex form by completing the square:

\(\begin{align*} g(x)&=x^2+6x+4\\ &=x^2+6x+9-9+4\\ &=(x+3)^2-5\\ \end{align*}\)

The vertex of \(g(x)\) is \((-3,-5)\).

Therefore, either \(D=\{x\in\mathbb{R} \mid x\ge-3\}\) or \(D=\{x\in\mathbb{R} \mid x\le-3\}\) is a restricted domain that ensures the inverse is a function and contains one complete arm of the parabola.

Check Your Understanding 3

Question — Version 1

Restrict the domain of \(f(x) = 2(x+4)(x-6)\) so that the inverse is a function and contains one complete arm of the parabola.

Domain is \(D=\{x \in \mathbb{R} \mid x \le a \}\) or \(D=\{x \in \mathbb{R} \mid x\ge b\}\).

Answer — Version 1

The two possible domain restrictions are \(D=\{x\in\mathbb{R} \mid x\ge 1\}\) or \(D=\{x\in\mathbb{R} \mid x\le 1\}\).

Feedback — Version 1

Since \(f(x) = 2(x+4)(x-6)\) is in factored form, we observe that:

• The zeros are  \( -4 \) and \(6\).
• The axis of symmetry, which lies halfway between the zeros, is \(x=1\).
• The \(x\)-coordinate of the vertex is \(1\) since the vertex lies on the axis of symmetry.

Therefore, either \(D=\{x\in\mathbb{R} \mid x\ge 1\}\) or \(D=\{x\in\mathbb{R} \mid x\le 1\}\) are restricted domains that ensure the inverse is a function and contains one complete arm of the parabola.

Question — Version 2

Restrict the domain of \(f(x) = -3(x+2)^2 - 3\) so that the inverse is a function and contains one complete arm of the parabola. 

Domain is \(D=\{x \in \mathbb{R} \mid x \le a \}\) or \(D=\{x \in \mathbb{R} \mid x\ge b\}\).

Answer — Version 2

The two possible domain restrictions are \(D=\{x\in\mathbb{R} \mid x\ge -2\}\) or \(D=\{x\in\mathbb{R} \mid x\le -2\}\).

Feedback — Version 2

Since \(f(x) = -3(x+2)^2 - 3\) is in vertex form, we observe that:

• The vertex is \((-2, -3)\).
• Specifically, the \(x\)-coordinate of the vertex is \(-2\).

Therefore, either \(D=\{x\in\mathbb{R} \mid x\ge -2\}\) or \(D=\{x\in\mathbb{R} \mid x\le -2\}\) are restricted domains that ensure the inverse is a function and contains one complete arm of the parabola.

Question — Version 3

Restrict the domain of \(f(x) = 2x^2 + 16x -2\) so that the inverse is a function and contains one complete arm of the parabola. 

Domain is \(D=\{x \in \mathbb{R} \mid x \le a \}\) or \(D=\{x \in \mathbb{R} \mid x\ge b\}\).

Answer — Version 3

The two possible domain restrictions are \(D=\{x\in\mathbb{R} \mid x\ge -4\}\) or \(D=\{x\in\mathbb{R} \mid x\le -4\}\).

Feedback — Version 3

Since \(f(x) = 2x^2 + 16x -2\) is in standard form, we first convert it to vertex form by completing the square:

\(\begin{align*} f(x)&= 2x^2 + 16x -2\\ &=2(x^2 + 8x) -2 \\ &=2(x^2 +8x +16 -16) - 2\\ &=2[(x+4)^2 - 16] -2\\ &=2(x+4)^2 - 32-2\\ &=2(x+4)^2 - 34 \end{align*}\)

The vertex of \(f(x)\) is \((-4, -34)\).

Therefore, either \(D=\{x\in\mathbb{R} \mid x\ge -4\}\) or \(D=\{x\in\mathbb{R} \mid x\le -4\}\) are restricted domains that ensure the inverse is a function and contains one complete arm of the parabola.

Wrap-Up

Take It With You

\(f(x)=\sin(x)\) is a function, but its inverse is not a function. In other words, \(f(x)\) is a many-to-one function.

See if you can find a way to restrict the domain of \(f(x)\) such that

• key features of the graph of \(y=f(x)\) are retained, yet
• the inverse is a function.

If you are not already familiar with the graph of \(y=\sin(x)\), you will likely need to use graphing technology to begin this question.