Explainer: Graphing Cubic Functions

In this explainer, we will learn how to graph cubic functions and write their rule from the given graph.

The simplest cubic function is the function that gives the volume of a cube of side π‘₯: 𝑓(π‘₯)=π‘₯ for positive π‘₯.

Let’s extend this function to all π‘₯ values and make a table of values to graph it.

π‘₯βˆ’3βˆ’2βˆ’10123
𝑦=π‘₯οŠ©βˆ’27βˆ’8βˆ’101827

The main characteristics of the cubic function are the following:

  • The function’s value is negative when π‘₯ is negative, positive when π‘₯ is positive, and 0 when π‘₯=0.
  • It is an odd function: 𝑓(π‘₯)=βˆ’π‘“(π‘₯), and, as such, its graph is symmetrical with respect to the origin.

Now, we are going to see how the graph changes when we slightly change the function to 𝑔(π‘₯)=π‘Žπ‘₯.

Let’s look at the graph of 𝑔(π‘₯)=βˆ’π‘₯. We see that 𝑔(π‘₯) and 𝑓(π‘₯) are always of opposite signs. We therefore expect the graph of 𝑔(π‘₯) to be symmetrical to the graph of 𝑓(π‘₯) with respect to the π‘₯-axis.

Second, let’s look at the graphs of the functions of the general equation β„Ž(π‘₯)=π‘Žπ‘₯ with π‘Žβ‰ 1; for instance, β„Ž(π‘₯)=2π‘₯ and β„Ž(π‘₯)=12π‘₯.

We can easily observe in the diagram the effect of the coefficient in front of π‘₯: if it is greater than 1, then the graph is steeper; if it is less than 1, then the graph is flatter than the graph of 𝑓(π‘₯)=π‘₯. If you work out the values of the three functions for π‘₯=1, you can easily see why the graph of β„Ž(π‘₯)=2π‘₯ is steeper and that of β„Ž(π‘₯)=12π‘₯ is flatter: for any π‘₯, their absolute values are, respectively, bigger and smaller.

𝑓(1)1
β„Ž(1)2
β„Ž(1)0.5

And, finally, let’s look at the graphs of the functions of the general equation 𝑙(π‘₯)=π‘₯+π‘˜οŠ©; for instance, 𝑙(π‘₯)=π‘₯+2 and 𝑙(π‘₯)=π‘₯βˆ’1.

We observe that 𝑙(π‘₯)=π‘₯+2 and 𝑙(π‘₯)=π‘₯βˆ’1 do not go through the origin. We see that 𝑙(π‘₯) is simply 𝑓(π‘₯)=π‘₯ translated up by two units. Indeed, for any value of π‘₯, 𝑙(π‘₯)=𝑓(π‘₯)+2, so (π‘₯,𝑙(π‘₯)) is two units above (π‘₯,𝑓(π‘₯)). And 𝑙(π‘₯) is 𝑓(π‘₯) translated downwards by one unit.

Let’s look now at one example.

Example 1: Matching a Graph with a Cubic Function

Which graph represents the function 𝑦=βˆ’2.5π‘₯+3?

Answer

We need to identify which of the given graphs represents the function 𝑦=βˆ’2.5π‘₯+3. The cubic function is in the form π‘Žπ‘₯+π‘˜οŠ©.

  • The absolute value of π‘Ž is greater than 1, so we know that the graph will be steeper than the graph of 𝑦=π‘₯.
  • π‘Ž is negative, so the function’s values are positive for π‘₯<0 and negative for π‘₯>0. Hence, answers (a) and (b) can be eliminated.
  • The value of π‘˜ is 3, so the graph of 𝑦=βˆ’2.5π‘₯ has been translated up 3 units. The answer (d) can be eliminated.

The answer must be (c). We can check the value of the function for π‘₯=1; we find 𝑦=βˆ’2.5β‹…1+3=0.5. The point (1,0.5) is indeed a point of the curve displayed in (c).

The answer is (c).

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