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 : for positive .

Letβs extend this function to all values and make a table of values to graph it.

0 | 1 | 2 | 3 | ||||

0 | 1 | 8 | 27 |

The main characteristics of the cubic function are the following:

- The functionβs value is negative when is negative, positive when is positive, and 0 when .
- It is an odd function: , 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 .

Letβs look at the graph of . We see that and are always of opposite signs. We therefore expect the graph of to be symmetrical to the graph of with respect to the -axis.

Second, letβs look at the graphs of the functions of the general equation with ; for instance, and .

We can easily observe in the diagram the effect of the coefficient in front of : 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 . If you work out the values of the three functions for , you can easily see why the graph of is steeper and that of is flatter: for any , their absolute values are, respectively, bigger and smaller.

1 | |

2 | |

0.5 |

And, finally, letβs look at the graphs of the functions of the general equation ; for instance, and .

We observe that and do not go through the origin. We see that is simply translated up by two units. Indeed, for any value of , , so is two units above . And is translated downwards by one unit.

Letβs look now at one example.

### Example 1: Matching a Graph with a Cubic Function

Which graph represents the function ?

### Answer

We need to identify which of the given graphs represents the function . The cubic function is in the form .

- The absolute value of is greater than 1, so we know that the graph will be steeper than the graph of .
- is negative, so the functionβs values are positive for and negative for . Hence, answers (a) and (b) can be eliminated.
- The value of is 3, so the graph of 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 ; we find . The point is indeed a point of the curve displayed in (c).

The answer is (c).