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Lesson Explainer: Graphing Factorized Cubic Functions Mathematics

In this explainer, we will learn how to graph cubic functions written in factored form and identify where they cross the axes.

A cubic function is one of the form 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+𝑐π‘₯+π‘‘οŠ©οŠ¨, where π‘Ž, 𝑏, 𝑐, and 𝑑 are real numbers and π‘Ž is nonzero. In order to sketch the graph of 𝑦=𝑓(π‘₯), we first recall that the coordinates of any point on the graph of this curve will have coordinates in the form (π‘₯,𝑓(π‘₯)). In other words, the π‘₯-coordinate tells us the input value of the function and the 𝑦-coordinate tells us the corresponding output.

It is difficult to sketch the graph of cubic functions in general since they can have many shapes. For example, consider the following graphs of 𝑦=π‘₯ and 𝑦=βˆ’π‘₯.

These graphs have similar shapes; however, in the first graph, as the values of π‘₯ increase, the outputs of the function are increasing without bound, so the curve keeps increasing upward. In the second graph, as the values of π‘₯ increase, the outputs of the function are decreasing without bound, so the curve keeps decreasing downward.

The description of the curve as the values of π‘₯ increase (or decrease) without bound is called the end behavior of the graphs. We sometimes refer to this in terms of π‘₯ approaching positive (or negative) infinity. Therefore, by looking at the graph of 𝑦=π‘₯, we can say that as the values of π‘₯ approach infinity, the outputs also approach infinity. However, when the values of π‘₯ approach negative infinity, the outputs also approach negative infinity.

This is not the only difference we can find in the shapes of different cubic functions. Consider the following graph of 𝑦=π‘₯(π‘₯+1)(π‘₯+2).

The end behavior of this graph is the same as 𝑦=π‘₯; however, the shape is different. We can see that there are three distinct π‘₯-intercepts and the curve turns twice.

We can find the coordinates of the π‘₯-intercepts by noting that the 𝑦-coordinates of these points must be 0. Therefore, the function outputs 0 at these points; they are the roots of the function.

In general, finding the π‘₯-intercepts of a cubic graph means solving the corresponding equation to find the roots, which is not something we have covered yet. However, in the case of factored cubic equations, we can determine the roots of a factored polynomial by noting that for a product to be equal to 0, one of the factors must be equal to 0. For example, the roots of the cubic π‘₯(π‘₯+1)(π‘₯+2) will be the solutions to the equation π‘₯(π‘₯+1)(π‘₯+2)=0.

We see that either π‘₯=0, π‘₯+1=0, or π‘₯+2=0. Solving each equation gives us π‘₯=βˆ’2, βˆ’1, or 0. We can then see that these are the π‘₯-intercepts of the graph of this function.

Property: π‘₯-Intercepts of a Factored Cubic Polynomial

The π‘₯-intercepts of the graph of 𝑦=π‘Ž(π‘₯βˆ’π‘)(π‘₯βˆ’π‘)(π‘₯βˆ’π‘‘) are at π‘₯=𝑏, π‘₯=𝑐, and π‘₯=𝑑.

In general, all cubic polynomials have at most three real roots, so their graphs have at most three π‘₯-intercepts.

When π‘₯ becomes very large, all the factors π‘₯βˆ’π‘, π‘₯βˆ’π‘, and π‘₯βˆ’π‘‘ will be positive, so their product will be positive. Consequently, if π‘Ž is positive too, then 𝑦 will be positive for large values of π‘₯. On the other hand, if π‘Ž is negative, then 𝑦 will be negative for large values of π‘₯.

Similarly, if we consider large negative values of π‘₯, we can see that all factors π‘₯βˆ’π‘, π‘₯βˆ’π‘, and π‘₯βˆ’π‘‘ will be negative, so their product will be negative too. Thus, if π‘Ž is positive, then 𝑦 will be negative for large negative values of π‘₯, and if π‘Ž is negative, then 𝑦 will be positive.

We note that this observation also applies to the nonfactored form of a cubic function. That is, the leading coefficient (i.e., the π‘Ž in π‘Žπ‘₯+𝑏π‘₯+𝑐π‘₯+π‘‘οŠ©οŠ¨) will determine the end behavior of the graph. This line of reasoning leads us to the following property.

Property: End Behavior of the Graphs of Cubic Functions

The end behavior of the graphs of cubic functions is determined by the sign of their leading coefficients.

If the leading coefficient is positive, then as π‘₯ increases without bound, the outputs increase without bound. Similarly, as π‘₯ decreases without bound, the outputs decrease without bound.

If the leading coefficient is negative, then this behavior is reversed; as π‘₯ increases without bound, the outputs decrease without bound, and as π‘₯ decreases without bound, the outputs increase without bound.

Cubic graphs with positive leading coefficients have the same end behavior as 𝑦=π‘₯.

Cubic graphs with negative leading coefficients have the same end behavior as 𝑦=βˆ’π‘₯.

One final piece of information we can use to sketch the graph of a cubic function is the location of the 𝑦-intercept. Recall that the 𝑦-intercept is where the curve intersects the 𝑦-axis, which happens when π‘₯=0. This is straightforward for us to calculate since we just need to substitute π‘₯=0 into the function (which can be done regardless of whether the function is in factored form).

We sketch the graph of a cubic polynomial by factoring, finding its π‘₯- and 𝑦-intercepts, and then checking the sign of its leading coefficient to determine the end behavior.

Let’s now see an example of using this method to sketch the graph of a cubic function.

Example 1: Identifying the Graph of a Factored Cubic Function

Which of the following is the graph of 𝑓(π‘₯)=(π‘₯βˆ’3)(π‘₯+1)(π‘₯+4)?

Answer

The easiest way to answer this question is to eliminate options. We can do this by recalling that the π‘₯-intercepts occur when the function outputs 0. This means that we can find the π‘₯-intercepts by solving the equation 𝑓(π‘₯)=0. This gives us (π‘₯βˆ’3)(π‘₯+1)(π‘₯+4)=0.

For a product to be equal to 0, one of the factors must be 0. We can set each factor equal to 0 and solve to determine the π‘₯-intercepts. We have π‘₯βˆ’3=0,π‘₯=3,π‘₯+1=0,π‘₯=βˆ’1,π‘₯+4=0,π‘₯=βˆ’4.whichgivesuswhichgivesuswhichgivesus

Therefore, there are three π‘₯-intercepts of the curve at π‘₯=βˆ’4, βˆ’1, and 3. We can see that only options C and D have these π‘₯-intercepts, so these are the only possible correct graphs.

There are now two possible ways we can determine the correct graph.

We can determine the 𝑦-intercept by substituting π‘₯=0 into the function. This gives us 𝑓(0)=(0βˆ’3)(0+1)(0+4)=(βˆ’3)(1)(4)=βˆ’12.

We see that the graph in option C has a positive 𝑦-intercept, so this cannot be the correct graph. This only leaves the graph in option D.

Alternatively, we can note that the leading coefficient is 1, and since this is positive, the end behavior of the graph will be the same as 𝑦=π‘₯. This means that the graph of the function should approach infinity as π‘₯ approaches infinity. We can see that the graph in option C does not have this end behavior. So, the answer must be option D.

It is worth noting that in the above example, we actually found enough information to sketch the graph of the function ourselves. To do this, we start by marking the three π‘₯-intercepts and 𝑦-intercepts as shown.

We then note that the end behavior of the graph is the same as 𝑦=π‘₯, so we can connect the points with a cubic curve that goes to infinity as π‘₯ approaches infinity to get the following graph.

In our next example, we will identify the correct equation of a cubic function from its graph.

Example 2: Identifying a Factored Cubic Function from Its Graph

Which equation matches the graph?

  1. 𝑓(π‘₯)=(π‘₯βˆ’1)(π‘₯+2)(π‘₯+3)
  2. 𝑓(π‘₯)=0.25(π‘₯βˆ’1)(π‘₯+2)(π‘₯+3)
  3. 𝑓(π‘₯)=βˆ’0.25(π‘₯βˆ’1)(π‘₯+2)(π‘₯+3)
  4. 𝑓(π‘₯)=βˆ’(π‘₯βˆ’1)(π‘₯+2)(π‘₯+3)
  5. 𝑓(π‘₯)=0.25(π‘₯+1)(π‘₯βˆ’2)(π‘₯βˆ’3)

Answer

We first note that all five of the given choices are factored cubic functions. We can recall that the π‘₯-intercepts occur when the function outputs 0, and for a factored function, they occur when each factor is equal to 0. We can note that we can read off the π‘₯-intercepts from the graph; they are at π‘₯=βˆ’3, π‘₯=βˆ’2, and π‘₯=1.

We can use these values to determine the possible linear factors of the function. Since the function outputs 0 at these values and the function is a product of factors, one of the factors must be equal to 0 at each π‘₯-intercept. This means that we have factors of (π‘₯βˆ’1), (π‘₯+2), and (π‘₯+3).

This tells us that our function must be of the form 𝑓(π‘₯)=π‘Ž(π‘₯βˆ’1)(π‘₯+2)(π‘₯+3), where π‘Ž is some real value. In order to determine the value of π‘Ž, we can examine the value of 𝑓(π‘₯) at a nonzero point, such as the 𝑦-intercept.

The 𝑦-intercept can be read off the graph by seeing where it intercepts the 𝑦-axis, which is at βˆ’1.5. This means that 𝑓(0)=βˆ’1.5. Substituting π‘₯=0 into our expression for 𝑓(π‘₯) gives us βˆ’1.5=π‘Ž(0βˆ’1)(0+2)(0+3)βˆ’1.5=βˆ’6π‘Ž.

Dividing through by βˆ’6 then gives us π‘Ž=βˆ’1.5βˆ’6=0.25.

This gives us the equation 𝑓(π‘₯)=0.25(π‘₯βˆ’1)(π‘₯+2)(π‘₯+3), which is option B.

Before we move on to our next example, there is one final possibility we need to consider, that is, what happens when we have repeated roots. For example, imagine we wanted to sketch the graph of the cubic function 𝑓(π‘₯)=(π‘₯βˆ’2)(π‘₯+1).

We can see that the function only has two roots, so there are only two π‘₯-intercepts of its graph. These are at π‘₯=βˆ’1 and π‘₯=2. We can find its 𝑦-intercept by evaluating 𝑓(0). We have 𝑓(0)=(0βˆ’2)(0+1)=βˆ’2.

We could then plot points to get an idea of the shape of the graph. If we did, we would get the following.

We can see that there appears to be two points where the graph turns and that the end behavior of the graph matches that of 𝑦=π‘₯. We can also notice that the graph only touches the π‘₯-axis at the point (βˆ’1,0) rather than intersecting it since there is no other intercept close to this point.

In general, if we have a cubic function 𝑓(π‘₯)=π‘Ž(π‘₯+𝑝)(π‘₯+π‘ž), where π‘β‰ π‘ž, then the graph will only touch the π‘₯-axis at the point (βˆ’π‘ž,0).

There is one final possibility, which is that the same root is repeated three times. For example, the function 𝑓(π‘₯)=(π‘₯βˆ’2). In this case, we just have a single π‘₯-intercept at π‘₯=2, and the end behavior is determined by the sign of the leading coefficient. In cases with three repeated roots, the shape of the curve is similar to 𝑦=π‘₯ (since this also has three repeated roots).

Let’s now see an example of using this process to identify the correct graph of a cubic function with a repeated factor.

Example 3: Identifying the Graph of a Factored Cubic Function with a Repeated Factor

Which of the following graphs represents 𝑓(π‘₯)=βˆ’(π‘₯+1)(π‘₯βˆ’2)?

Answer

We note that we are asked to determine the correct graph of a factored cubic polynomial. We can start by finding the π‘₯-intercepts and 𝑦-intercepts. Let’s start with the π‘₯-intercepts. These occur when the outputs of the function are 0. This function is the product of factors, so one of the factors must be equal to 0 at the π‘₯-intercepts. We have π‘₯+1=0,π‘₯=βˆ’1,π‘₯βˆ’2=0,π‘₯=2.whichgivesuswhichgivesus

We can find the 𝑦-intercept by substituting π‘₯=0 into the function. We calculate 𝑓(0)=βˆ’(0+1)(0βˆ’2)=βˆ’(1)(βˆ’2)=βˆ’4.

At this point, we can see that only option A has a 𝑦-intercept of βˆ’4, so it is the only possible correct graph of the function. However, it is always good practice to find all the information so that we can sketch our own graph of the function.

We can also see that the function has a repeated factor of (π‘₯βˆ’2). This means that the function will only touch the π‘₯-axis at this point. Finally, we can note that the leading coefficient is negative, so the end behavior of the graph will be the same as that of 𝑦=βˆ’π‘₯.

We can add the π‘₯-intercepts and 𝑦-intercepts to a graph and note the end behavior.

Sketching a cubic-shaped curve through these points making sure to only touch the point (2,0) gives us the following.

We see that this sketch matches the graph in option A.

In our next example, we will find the values of three missing constants from a cubic function by using its graph.

Example 4: Finding the Unknown Variables in a Cubic Function given Its Graph

The following graph represents the function 𝑓(π‘₯)=π‘˜(π‘₯βˆ’π‘Ž)(π‘₯βˆ’π‘). Find the values of π‘˜, π‘Ž, and 𝑏.

Answer

We are given the graph of a cubic function and are asked to determine the values of three constants from the function. We can do this by recalling the relationships between properties of the graph and the structure of the function.

First, we recall that the π‘₯-intercepts of a factored cubic function occur when each factor is equal to 0. In particular, if (π‘₯βˆ’π‘) is a factor, the graph will have an π‘₯-intercept at 𝑝. We can see in the diagram that the graph has two π‘₯-intercepts: βˆ’1 and 2. This means that (π‘₯βˆ’(βˆ’1))=(π‘₯+1) and (π‘₯βˆ’2) are factors of the cubic function.

In order to find which factor is repeated (i.e., has a square), we recall that repeated factors result in an π‘₯-intercept of the graph that only touches the π‘₯-axis at a point. Since this occurs at π‘₯=βˆ’1, this means that the factor of (π‘₯+1) must be a repeated factor. This gives us 𝑓(π‘₯)=π‘˜(π‘₯βˆ’2)(π‘₯+1).

Thus, π‘Ž=2 and 𝑏=βˆ’1.

To determine the value of π‘˜, we can use the 𝑦-intercept of the graph. From the diagram, we can see that the 𝑦-intercept is at 4. This means that 𝑓(0)=4. We can substitute π‘₯=0 into our function to get 𝑓(0)=π‘˜(0βˆ’2)(0+1)=π‘˜(βˆ’2)(1)=βˆ’2π‘˜.

Since 𝑓(0)=4,this tells us that 4=βˆ’2π‘˜. Thus, π‘˜=4βˆ’2=βˆ’2.

This gives us 𝑓(π‘₯)=βˆ’2(π‘₯βˆ’2)(π‘₯+1).

Hence, π‘˜=βˆ’2, π‘Ž=2, and 𝑏=βˆ’1.

In general, it is very difficult to factor a cubic polynomial. Factoring cubics often requires advanced methods; however, there are examples of easier ways to factor cubics. For example, if all the terms in the cubic share a factor of π‘₯, then we can take out this factor, leaving us with a quadratic. We can then apply our tools for factoring quadratics.

In our next example, we will factor a cubic polynomial to help us determine the correct sketch of its curve.

Example 5: Factoring and Identifying the Graph of a Cubic Function

  1. Factor the cubic polynomial π‘₯βˆ’π‘₯.
  2. Determine which of the following is the correct sketch of the curve 𝑦=π‘₯βˆ’π‘₯.

Answer

Part 1

We want to factor the cubic polynomial π‘₯βˆ’π‘₯. We can do this by noting that both terms share a factor of π‘₯. Taking out the shared factor of π‘₯ gives us π‘₯βˆ’π‘₯=π‘₯ο€Ήπ‘₯βˆ’1.

We can then note that π‘₯βˆ’1 is a difference between two squares. This means that we can factor this as π‘₯βˆ’1=(π‘₯βˆ’1)(π‘₯+1).

Substituting this into the function gives us π‘₯βˆ’π‘₯=π‘₯ο€Ήπ‘₯βˆ’1=π‘₯(π‘₯βˆ’1)(π‘₯+1).

Part 2

We can determine the π‘₯-intercepts of the graph of this function by setting each factor equal to 0 and solving for π‘₯. We find that the π‘₯-intercepts are at π‘₯=0, π‘₯=1, and π‘₯=βˆ’1. This means that choices A, B, and E cannot be correct since they do not have these π‘₯-intercepts.

We can find the 𝑦-intercept of the curve by substituting π‘₯=0 into the function. This yields 0(0βˆ’1)(0+1)=0.

So, the 𝑦-intercept is at 0, which is consistent with the fact that π‘₯=0 is an π‘₯-intercept.

Finally, we can determine the end behavior of the curve by noting that the leading coefficient of the polynomial is positive. This means that the end behavior will be the same as 𝑦=π‘₯. In particular, as the values of π‘₯ increase without bound, the outputs of the function should increase without bound. This is not true in choice C, so the correct answer must be choice D.

In our final example, we will factor a cubic polynomial to help us determine the sketch of a graph.

Example 6: Factoring and Identifying the Graph of a Cubic Function

  1. Factor the cubic polynomial 2π‘₯+2π‘₯βˆ’4π‘₯.
  2. Determine which of the following is the correct sketch of the curve 𝑦=2π‘₯+2π‘₯βˆ’4π‘₯.

Answer

Part 1

We want to factor the cubic polynomial 2π‘₯+2π‘₯βˆ’4π‘₯. We can do this by noting that all the terms share a factor of 2π‘₯. Taking out the shared factor of π‘₯ gives us 2π‘₯+2π‘₯βˆ’4π‘₯=2π‘₯ο€Ήπ‘₯+π‘₯βˆ’2.

To factor the quadratic, we need to find two numbers whose product is βˆ’2 and that add to 1. We can find that 2Γ—(βˆ’1)=βˆ’2 and that 2+(βˆ’1)=1. This allows us to factor the quadratic as follows: π‘₯+π‘₯βˆ’2=(π‘₯+2)(π‘₯βˆ’1).

Substituting this into the cubic polynomial gives 2π‘₯+2π‘₯βˆ’4π‘₯=2π‘₯ο€Ήπ‘₯+π‘₯βˆ’2=2π‘₯(π‘₯+2)(π‘₯βˆ’1).

Part 2

We note that we are asked to determine the correct graph of a factored cubic polynomial. We can start by finding the π‘₯-intercepts and 𝑦-intercepts. Let’s start with the π‘₯-intercepts. These occur when the output of the function is 0. As we have factored the function, we can find the π‘₯-intercepts by setting each of the factors to be zero in turn. We have π‘₯=0,π‘₯βˆ’1=0,π‘₯=1,π‘₯+2=0,π‘₯=βˆ’2.whichgivesuswhichgivesus

Thus, our graph has three distinct π‘₯-intercepts at βˆ’2, 0, and 1. As one of the π‘₯-intercepts is at (0,0), this also gives us the 𝑦-intercept of the graph. We can see that only choices B and C have these π‘₯-intercepts, so it must be one of these graphs.

Finally, we can note that the leading coefficient is positive, so the end behavior of the graph will be the same as that of 𝑦=π‘₯.

We can add the π‘₯-intercepts and 𝑦-intercepts onto a graph and note the end behavior.

Sketching a cubic-shaped curve through these points with the correct end behavior gives us the following.

We can see that this sketch matches the graph given in choice B.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • The π‘₯-intercepts of the graph of 𝑦=π‘Ž(π‘₯βˆ’π‘)(π‘₯βˆ’π‘)(π‘₯βˆ’π‘‘) are at π‘₯=𝑏, π‘₯=𝑐, and π‘₯=𝑑.
  • All cubic polynomials have at most three real roots, so their graphs have at most three π‘₯-intercepts.
  • The end behavior of the graphs of cubic functions is determined by the sign of their leading coefficients. If the leading coefficient is positive, then the cubic function has the same end behavior as π‘₯. If the leading coefficient is negative, then the cubic function has the same end behavior as βˆ’π‘₯.
  • If there is a repeated root, such as 𝑦=π‘Ž(π‘₯βˆ’π‘)(π‘₯βˆ’π‘ž), then the graph only touches the π‘₯-axis at π‘ž.

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