Lesson Explainer: Cubic Functions and Their Graphs Mathematics

In this explainer, we will learn how to graph cubic functions, write their rules from their graphs, and identify their features.

We will focus on the standard cubic function, 𝑓(π‘₯)=π‘₯. Creating a table of values with integer values of π‘₯ from βˆ’2≀π‘₯≀2, we can then graph the function.

π‘₯βˆ’2βˆ’1012
𝑓(π‘₯)=π‘₯οŠ©βˆ’8βˆ’1018

The main characteristics of the cubic function are the following:

  • The value of the function is positive when π‘₯ is positive, negative when π‘₯ is negative, and 0 when π‘₯=0.
  • As a function with an odd degree (3), it has opposite end behaviors. Its end behavior is such that as π‘₯ increases to infinity, 𝑓(π‘₯) also increases to infinity. As π‘₯ decreases, 𝑓(π‘₯) also decreases to negative infinity.
  • It is an odd function, 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯), for all values of π‘₯ in the domain of 𝑓, and, as such, its graph is invariant under a rotation of 180∘ about the origin.

When we transform this function, the definition of the curve is maintained. We will look at a number of different transformations, and we can consider these to be of two types:

  • Changes to the input, π‘₯, for example, π‘₯⟢π‘₯+3 or π‘₯⟢4π‘₯
  • Changes to the output, 𝑓(π‘₯), for example, 𝑓(π‘₯)βŸΆπ‘“(π‘₯)βˆ’3 or 𝑓(π‘₯)⟢2𝑓(π‘₯)

Furthermore, we can consider the changes to the input, π‘₯, and the output, 𝑓(π‘₯), as consisting of

  • addition,
  • multiplication,
  • negation.

We can now investigate how the graph of the function changes when we add or subtract values from the output.

Say we have the functions 𝑔(π‘₯) and β„Ž(π‘₯) such that 𝑔(π‘₯)=𝑓(π‘₯)+2 and β„Ž(π‘₯)=𝑓(π‘₯)βˆ’1, then 𝑓(π‘₯)=π‘₯,𝑔(π‘₯)=π‘₯+2,β„Ž(π‘₯)=π‘₯βˆ’1.

In order to plot the graphs of these functions, we can extend the table of values above to consider the values of 𝑓(π‘₯) for the same values of π‘₯. Thus, we have the table below.

π‘₯βˆ’2βˆ’1012
𝑓(π‘₯)=π‘₯οŠ©βˆ’8βˆ’1018
𝑔(π‘₯)=π‘₯+2οŠ©βˆ’612310
β„Ž(π‘₯)=π‘₯βˆ’1οŠ©βˆ’9βˆ’2βˆ’107

The outputs of 𝑔 are always 2 larger than those of 𝑓. Similarly, each of the outputs of β„Ž is 1 less than those of 𝑓. We can graph these three functions alongside one another as shown.

We observe that these functions are a vertical translation of 𝑓(π‘₯)=π‘₯.

In general, the graph of a function 𝑓(π‘₯)+π‘˜, for a constant π‘˜βˆˆβ„, is a vertical translation of the graph of the function 𝑓(π‘₯)=π‘₯. If π‘˜>0, then its graph is a translation π‘˜ units upward of the graph of 𝑓(π‘₯)=π‘₯. If π‘˜<0, then its graph is a translation of |π‘˜| units downward of the graph of 𝑓(π‘₯)=π‘₯.

Next, we can investigate how the function 𝑓(π‘₯)=π‘₯ changes when we add values to the input. This time, we take the functions 𝑔(π‘₯) and β„Ž(π‘₯) such that 𝑔(π‘₯)=𝑓(π‘₯+2) and β„Ž(π‘₯)=𝑓(π‘₯βˆ’1): 𝑓(π‘₯)=π‘₯,𝑔(π‘₯)=(π‘₯+2),β„Ž(π‘₯)=(π‘₯βˆ’1).

We can create a table of values for these functions and plot a graph of these functions.

π‘₯βˆ’2βˆ’1012
𝑓(π‘₯)=π‘₯οŠ©βˆ’8βˆ’1018
𝑔(π‘₯)=(π‘₯+2)0182764
β„Ž(π‘₯)=(π‘₯βˆ’1)οŠ©βˆ’27βˆ’8βˆ’101

We observe that the graph of the function 𝑔(π‘₯)=(π‘₯+2) is a horizontal translation of 𝑓(π‘₯)=π‘₯ two units left. Similarly, β„Ž(π‘₯)=(π‘₯βˆ’1) is a horizontal translation of 𝑓(π‘₯)=π‘₯ one unit right.

This can be a counterintuitive transformation to recall, as we often consider addition in a translation as producing a movement in the positive direction. In this case, the reverse is true. In order to help recall this property, we consider that the function is translated horizontally β„Ž units right by a change to the input, π‘₯⟢π‘₯βˆ’β„Ž. Therefore, for example, in the function 𝑓(π‘₯)=(π‘₯+1), β„Ž=βˆ’1, and the function 𝑓(π‘₯)=π‘₯ is translated left 1 unit.

We can summarize how addition changes the function 𝑓(π‘₯)=π‘₯ below.

OperationTransformed EquationGeometric Change
𝑓(π‘₯)βŸΆπ‘“(π‘₯)+π‘˜π‘“(π‘₯)=π‘₯+π‘˜οŠ©Vertical translation:
If π‘˜>0, π‘˜ units up.
If π‘˜<0, |π‘˜| units down.
π‘₯⟢π‘₯βˆ’β„Žπ‘“(π‘₯)=(π‘₯βˆ’β„Ž)Horizontal translation:
If β„Ž>0, β„Ž units right.
If β„Ž<0, |β„Ž| units left.

Next, we can investigate how multiplication changes the function 𝑓(π‘₯)=π‘₯, beginning with changes to the output, 𝑓(π‘₯).

Let us consider the functions 𝑓(π‘₯)=π‘₯, 𝑔(π‘₯)=3𝑓(π‘₯), and β„Ž(π‘₯)=12𝑓(π‘₯): 𝑓(π‘₯)=π‘₯,𝑔(π‘₯)=3π‘₯,β„Ž(π‘₯)=12π‘₯.

We can observe that the function 𝑓(π‘₯)=3π‘₯ has been stretched vertically, or dilated, by a factor of 3. The function 𝑓(π‘₯)=12π‘₯ has a vertical dilation by a factor of 12. Thus, for any positive value of π‘Ž when 𝑓(π‘₯)βŸΆπ‘Žπ‘“(π‘₯), there is a vertical stretch of factor π‘Ž.

If we change the input, π‘₯βŸΆπ‘π‘₯, for 𝑏>0, we would have a function of the form β„Ž(π‘₯)=(𝑏π‘₯). The following graph compares the function 𝑓(π‘₯)=π‘₯ with β„Ž(π‘₯)=(2π‘₯).

If we consider the coordinates (1,1) in the function 𝑓(π‘₯)=π‘₯, we will find that this is when the input, 1, produces an output of 1. To get the same output value of 1 in the function β„Ž(π‘₯)=(2π‘₯), 2π‘₯=1; so π‘₯=12. The same is true for the coordinates (2,8) in 𝑓(π‘₯)=π‘₯. The same output of 8 in β„Ž(π‘₯)=(2π‘₯) is obtained when 2π‘₯=2, so π‘₯=1. Thus, when we multiply every π‘₯ value in 𝑓(π‘₯) by 2, to obtain the function β„Ž(π‘₯)=(2π‘₯), the graph of 𝑓(π‘₯)=π‘₯ is dilated horizontally by a factor of 12, with each point being moved to one-half of its previous distance from the 𝑦-axis.

For any positive π‘βˆˆβ„ when π‘₯βŸΆπ‘π‘₯, the graph of 𝑦=(𝑏π‘₯) is a horizontal dilation of 𝑓(π‘₯)=π‘₯ by a factor of 1𝑏.

We can summarize these results below, for a positive π‘Ž and 𝑏.

OperationTransformed EquationGeometric Change
𝑓(π‘₯)βŸΆπ‘Žπ‘“(π‘₯)𝑓(π‘₯)=π‘Žπ‘₯Vertical dilation of factor π‘Ž
π‘₯βŸΆπ‘π‘₯𝑓(π‘₯)=(𝑏π‘₯)Horizontal dilation of factor 1𝑏

Finally, we can investigate changes to the standard cubic function by negation, for a function 𝑔(π‘₯)=βˆ’π‘“(π‘₯). This gives us the function 𝑔(π‘₯)=βˆ’π‘₯. Every output value of 𝑔(π‘₯) would be the negative of its value in 𝑓(π‘₯)=π‘₯. For example, the coordinates (2,8) in the original function would be (2,βˆ’8) in the transformed function.

This gives the effect of a reflection in the horizontal axis.

Since the cubic graph is an odd function, we know that 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯). Thus, changing the input π‘₯βŸΆβˆ’π‘₯ in the function 𝑓(π‘₯)=π‘₯ also transforms the function to 𝑓(π‘₯)=βˆ’π‘₯. In general, for any function 𝑓(π‘₯), 𝑓(π‘₯)βŸΆβˆ’π‘“(π‘₯) creates a reflection in the horizontal axis and changing the input π‘₯βŸΆβˆ’π‘₯ creates a reflection of 𝑓(π‘₯) in the vertical axis. The fact that the cubic function, 𝑓(π‘₯)=π‘₯, is odd means that negating either the input or the output produces the same graphical result.

OperationTransformed EquationGeometric Change
𝑓(π‘₯)βŸΆβˆ’π‘“(π‘₯)𝑓(π‘₯)=βˆ’π‘₯Reflection in the horizontal axis
π‘₯βŸΆβˆ’π‘₯𝑓(π‘₯)=(βˆ’π‘₯)Reflection in the vertical axis

We can create the complete table of changes to the function 𝑓(π‘₯)=π‘₯ below, for a positive π‘Ž and 𝑏.

OperationTransformed EquationGeometric Change
𝑓(π‘₯)βŸΆπ‘“(π‘₯)+π‘˜π‘“(π‘₯)=π‘₯+π‘˜οŠ©Vertical translation:
If π‘˜>0, π‘˜ units up.
If π‘˜<0, If π‘˜<0, |π‘˜| units down.
π‘₯⟢π‘₯βˆ’β„Žπ‘“(π‘₯)=(π‘₯βˆ’β„Ž)Horizontal translation:
If β„Ž>0, β„Ž units right.
If β„Ž<0, |β„Ž| units left.
𝑓(π‘₯)βŸΆπ‘Žπ‘“(π‘₯)𝑓(π‘₯)=π‘Žπ‘₯Vertical dilation of factor π‘Ž
π‘₯βŸΆπ‘π‘₯𝑓(π‘₯)=(𝑏π‘₯)Horizontal dilation of factor 1𝑏
𝑓(π‘₯)βŸΆβˆ’π‘“(π‘₯)𝑓(π‘₯)=βˆ’π‘₯Reflectioninthehorizontalaxis
π‘₯βŸΆβˆ’π‘₯𝑓(π‘₯)=(βˆ’π‘₯)Reflection in the vertical axis

We can combine a number of these different transformations to the standard cubic function, creating a function in the form 𝑓(π‘₯)=π‘Ž(π‘₯βˆ’β„Ž)+π‘˜οŠ©.

Definition: Transformations of the Cubic Function

If π‘Ž, β„Ž, and π‘˜βˆˆβ„, with π‘Žβ‰ 0, then the graph of 𝑓(π‘₯)=π‘Ž(π‘₯βˆ’β„Ž)+π‘˜οŠ© is a transformation of the graph of 𝑦=π‘₯. We list the transformations we need to transform the graph of 𝑦=π‘₯ into 𝑦=𝑓(π‘₯) as follows:

  • If π‘Ž>0, then the graph of 𝑦=π‘₯ is vertically dilated by a factor π‘Ž.
  • If π‘Ž<0, then the graph of 𝑦=π‘₯ is reflected in the horizontal axis and vertically dilated by a factor |π‘Ž|.
  • If β„Ž>0, then the graph of 𝑦=π‘₯ is translated horizontally β„Ž units right.
  • If β„Ž<0, then the graph of 𝑦=π‘₯ is translated horizontally |β„Ž| units left.
  • If π‘˜>0, then the graph of 𝑦=π‘₯ is translated vertically π‘˜ units up.
  • If π‘˜<0, then the graph of 𝑦=π‘₯ is translated vertically |π‘˜| units down.

The order in which we perform the transformations of a function is important, even if, on occasion, we obtain the same graph regardless. We use the following order:

  1. Vertical dilation, π‘Ž
  2. Horizontal translation, β„Ž
  3. Vertical translation, π‘˜

If we are given the graph of an unknown cubic function, we can use the shape of the parent function, 𝑓(π‘₯)=π‘₯, to establish which transformations have been applied to it and hence establish the function. Let us see an example of how we can do this.

Example 1: Writing the Equation of a Graph by Recognizing Translations of the Standard Cubic Function

Which equation matches the graph?

  1. 𝑦=(π‘₯βˆ’2)βˆ’1
  2. 𝑦=(π‘₯+2)βˆ’1
  3. 𝑦=(π‘₯+2)+1
  4. 𝑦=(π‘₯βˆ’2)+1

Answer

We may observe that this function looks similar in shape to the standard cubic function, 𝑓(π‘₯)=π‘₯, sometimes written as the equation 𝑦=π‘₯. In fact, we can note there is no dilation of the function, either by looking at its shape or by noting the coefficients of (π‘₯βˆ’β„Ž) in the given options are 1. The graph of 𝑦=π‘₯ passes through the origin and can be sketched on the same graph as shown below.

The inflection point of 𝑦=π‘₯ is at the coordinate (0,0), and the inflection point of the unknown function is at (βˆ’2,βˆ’1). Therefore, the function 𝑦=π‘₯ has been translated two units left and 1 unit down. As both functions have the same steepness and they have not been reflected, then there are no further transformations.

A cubic function in the form 𝑦=π‘Ž(π‘₯βˆ’β„Ž)+π‘˜οŠ© is a transformation of 𝑦=π‘₯, for π‘Ž, β„Ž, and π‘˜βˆˆβ„, with π‘Žβ‰ 0. In this form, the value of π‘Ž indicates the dilation scale factor, and a reflection if π‘Ž<0; there is a horizontal translation β„Ž units right and a vertical translation π‘˜ units up. We perform these transformations with the vertical stretch first, horizontal translation second, and vertical translation third.

In this question, the graph has not been reflected or dilated, so π‘Ž=1. And we do not need to perform any vertical dilation. Next, the function has a horizontal translation of 2 units left, so β„Ž=βˆ’2. The vertical translation of 1 unit down means that π‘˜=βˆ’1. We can fill these into the equation 𝑦=π‘Ž(π‘₯βˆ’β„Ž)+π‘˜οŠ©, which gives 𝑦=1(π‘₯βˆ’(βˆ’2))+(βˆ’1)=(π‘₯+2)βˆ’1.

Therefore, the equation of the graph is that given in option B: 𝑦=(π‘₯+2)βˆ’1.

In the following example, we will identify the correct shape of a graph of a cubic function.

Example 2: Identifying the Graph of a Cubic Function by Identifying Transformations of the Standard Cubic Function

Which of the following is the graph of 𝑓(π‘₯)=βˆ’(π‘₯βˆ’2)?

Answer

We can compare the function 𝑓(π‘₯)=βˆ’(π‘₯βˆ’2) with its parent function 𝑓(π‘₯)=π‘₯, which we can sketch below.

A cubic function in the form 𝑓(π‘₯)=π‘Ž(π‘₯βˆ’β„Ž)+π‘˜οŠ© is a transformation of 𝑓(π‘₯)=π‘₯, for π‘Ž, β„Ž, and π‘˜βˆˆβ„, with π‘Žβ‰ 0. Here, π‘Ž represents a dilation or reflection, β„Ž gives the number of units that the graph is translated in the horizontal direction, and π‘˜ is the number of units the graph is translated in the vertical direction. We perform these transformations with the vertical dilation first, horizontal translation second, and vertical translation third.

In the function 𝑓(π‘₯)=βˆ’(π‘₯βˆ’2), the value of π‘Ž=βˆ’1. This indicates that there is no dilation (or rather, a dilation of a scale factor of 1). However, since π‘Ž is negative, this means that there is a reflection of the graph in the π‘₯-axis. Hence, we could perform the reflection of 𝑓(π‘₯)=π‘₯ as shown below, creating the function 𝑓(π‘₯)=βˆ’π‘₯.

Next, in the given function, 𝑓(π‘₯)=βˆ’(π‘₯βˆ’2), the value of β„Ž is 2, indicating that there is a translation 2 units right. This moves the inflection point from (0,0) to (2,0). The function 𝑓(π‘₯)=βˆ’(π‘₯βˆ’2) could be sketched as shown.

Therefore, the graph that shows the function 𝑓(π‘₯)=βˆ’(π‘₯βˆ’2) is option E.

In the next example, we will see how we can write a function given its graph.

Example 3: Writing the Equation of a Graph by Recognizing Translations of the Standard Cubic Function

Select the equation of this curve.

  1. 𝑦=π‘₯βˆ’3
  2. 𝑦=βˆ’π‘₯βˆ’3
  3. 𝑦=2π‘₯+3
  4. 𝑦=βˆ’2π‘₯βˆ’3
  5. 𝑦=π‘₯+3

Answer

We can compare this function to the function 𝑦=π‘₯ by sketching the graph of this function on the same axes.

We note that there has been no dilation or reflection since the steepness and end behavior of the curves are identical. There is no horizontal translation, but there is a vertical translation of 3 units downward. For any value π‘˜βˆˆβ„, the function 𝑦=π‘₯+π‘˜οŠ© is a translation of the function 𝑦=π‘₯ by π‘˜ units vertically. As the translation here is in the negative direction, the value of π‘˜ must be negative; hence, π‘˜=βˆ’3.

Thus, the equation of this curve is the answer given in option A: 𝑦=π‘₯βˆ’3.

We will now see an example where we will need to identify three separate transformations of the standard cubic function.

Example 4: Identifying the Graph of a Cubic Function by Identifying Transformations of the Standard Cubic Function

Which of the following graphs represents 𝑓(π‘₯)=2βˆ’(π‘₯βˆ’5)?

Answer

A cubic function in the form 𝑦=π‘Ž(π‘₯βˆ’β„Ž)+π‘˜οŠ© is a transformation of 𝑦=π‘₯, for π‘Ž, β„Ž, and π‘˜βˆˆβ„, with π‘Žβ‰ 0. Here, π‘Ž represents a dilation or reflection, β„Ž gives the number of units that the graph is translated in the horizontal direction, and π‘˜ is the number of units the graph is translated in the vertical direction. We perform these transformations with the vertical stretch first, horizontal translation second, and vertical translation third.

The function 𝑓(π‘₯)=2βˆ’(π‘₯βˆ’5) can be written as 𝑓(π‘₯)=βˆ’(π‘₯βˆ’5)+2.

As the value π‘Ž=βˆ’1 is a negative value, the graph must be reflected in the π‘₯-axis. Since |π‘Ž|=1, the graph of 𝑦=π‘₯ has a vertical dilation of a scale factor of 1; thus, it will have the same shape.

As β„Ž=5, there is a horizontal translation of 5 units right.

Finally, π‘˜=2, so the graph also has a vertical translation of 2 units up.

We can visualize the translations in stages, beginning with the graph of 𝑓(π‘₯)=π‘₯.

Combining the two translations and the reflection gives us the solution that the graph that shows the function 𝑓(π‘₯)=2βˆ’(π‘₯βˆ’5) is option B.

As an aside, option A represents the function 𝑓(π‘₯)=2+(π‘₯+5), option C represents the function 𝑓(π‘₯)=2βˆ’(π‘₯+5), and option D is the function 𝑓(π‘₯)=2+(π‘₯βˆ’5).

We will now look at an example involving a dilation.

Example 5: Writing the Equation of a Graph by Recognizing Translations of the Standard Cubic Function

Which equation matches the graph?

  1. 𝑦=(π‘₯βˆ’1)+4
  2. 𝑦=2(π‘₯βˆ’1)+4
  3. 𝑦=3(π‘₯βˆ’1)+4
  4. 𝑦=13(π‘₯βˆ’1)+4
  5. 𝑦=12(π‘₯βˆ’1)+4

Answer

The function shown is a transformation of the graph of 𝑦=π‘₯. We can write the equation of the graph in the form 𝑦=π‘Ž(π‘₯βˆ’β„Ž)+π‘˜οŠ©, which is a transformation of 𝑦=π‘₯, for π‘Ž, β„Ž, and π‘˜βˆˆβ„, with π‘Žβ‰ 0. Here, π‘Ž represents a dilation or reflection, β„Ž gives the number of units that the graph is translated in the horizontal direction, and π‘˜ is the number of units the graph is translated in the vertical direction. We perform these transformations with the vertical stretch first, horizontal translation second, and vertical translation third.

We can sketch the graph of 𝑦=π‘₯ alongside the given curve.

If we compare the turning point of 𝑦=π‘₯ with that of the given graph, we have (0,0)⟢(1,4). This indicates a horizontal translation of 1 unit right and a vertical translation of 4 units up.

We observe that the given curve is steeper than that of the function 𝑦=π‘₯. We can compare a translation of 𝑦=π‘₯ by 1 unit right and 4 units up with the given curve.

An input, π‘₯, of 0 in the translated function produces an output, 𝑦, of 3. However, a similar input of 0 in the given curve produces an output of 1. There is a dilation of a scale factor of 3 between the two curves. As the given curve is steeper than that of the function 𝑓(π‘₯)=π‘₯, then it has been dilated vertically by a scale factor of 3 (rather than being dilated with a scale factor of 13, which would produce a β€œcompressed” graph).

We can now substitute π‘Ž=3, β„Ž=1, and π‘˜=4 into 𝑦=π‘Ž(π‘₯βˆ’β„Ž)+π‘˜οŠ© to give 𝑦=3(π‘₯βˆ’1)+4.

This is the answer given in option C.

We will look at a final example involving one of the features of a cubic function: the point of symmetry.

Example 6: Identifying the Point of Symmetry of a Cubic Function

Consider the graph of the function 𝑦=(π‘₯+2)βˆ’2.

Write down the coordinates of the point of symmetry of the graph, if it exists.

Answer

The given graph is a translation of 𝑦=π‘₯ by 2 units left and 2 units down. Since 𝑦=π‘₯ has a point of rotational symmetry at (0,0), then after a translation, the translated graph will have a point of rotational symmetry 2 units left and 2 units down from (0,0).

Therefore, we can identify the point of symmetry as (βˆ’2,βˆ’2).

We now summarize the key points.

Key Points

  • The standard cubic function is the function 𝑓(π‘₯)=π‘₯. It has the following properties:
    • The function’s outputs are positive when π‘₯ is positive, negative when π‘₯ is negative, and 0 when π‘₯=0.
    • Its end behavior is such that as π‘₯ increases to infinity, 𝑓(π‘₯) also increases to infinity. As π‘₯ decreases, 𝑓(π‘₯) also decreases to negative infinity.
    • It is an odd function, 𝑓(βˆ’π‘₯)=βˆ’π‘“(π‘₯), and, as such, its graph has 180∘ rotational symmetry about the origin.
  • If π‘Ž, β„Ž, and π‘˜βˆˆβ„, with π‘Žβ‰ 0, then the graph of 𝑓(π‘₯)=π‘Ž(π‘₯βˆ’β„Ž)+π‘˜οŠ© is a transformation of the graph of 𝑦=π‘₯. We list the transformations we need to transform the graph of 𝑦=π‘₯ into 𝑦=𝑓(π‘₯) as follows:
    • If π‘Ž>0, then the graph of 𝑦=π‘₯ is vertically dilated by a factor π‘Ž.
    • If π‘Ž<0, then the graph of 𝑦=π‘₯ is reflected in the horizontal axis and vertically dilated by a factor |π‘Ž|.
    • If β„Ž>0, then the graph of 𝑦=π‘₯ is translated horizontally β„Ž units right.
    • If β„Ž<0, then the graph of 𝑦=π‘₯ is translated horizontally |β„Ž| units left.
    • If π‘˜>0, then the graph of 𝑦=π‘₯ is translated vertically π‘˜ units up.
    • If π‘˜<0, then the graph of 𝑦=π‘₯ is translated vertically |π‘˜| units down.
  • For the function 𝑓(π‘₯)=π‘Ž(π‘₯βˆ’β„Ž)+π‘˜οŠ©, we perform transformations of the cubic function in the following order:
    • Vertical dilation, π‘Ž
    • Horizontal translation, β„Ž
    • Vertical translation, π‘˜

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