How to graph cubic equations?

A cubic equation in algebra in math is an equation in which the highest degree of the variable is 3. The general form of a cubic equation would be like this:

Y = ax^3 + bx^2 + cx + d



Here, ax^3 is the leading term. The leading coefficient is a. If this term is not there, or if the leading coefficient a  is equal to zero then the equation no longer remains a cubic equation and becomes a quadratic equation. Therefore every cubic equation would have the leading coefficient not equal to 0.

If the equation that we are asked to graph is not in the above form, then the first step would be to convert it to the above mentioned general form. That would make charting easier. For graphing cubic equations or for graphing any equation, the simplest method is to make a table of values for x and y. In that we assume various values of x and find the corresponding values of y. Then we plot these (x,y) ordered pairs as points on a graph sheet. Then we run a smooth curve through the points we plotted.  Let us now try to understand this process better using a sample problem question.

Sample problem 1:

Construct a cubic equation graph for the equation: 3x^2 + x + y = x^3 + 3


The first step would be to convert the given equation such that it resembles the standard form mentioned above. Therefore our equation becomes:

3x^2 + x + y = x^3 + 3 – Subtracting 3x^2 + x from both sides,

-3x^2 - x      = -3x^2 – x

Y = x^3 – 3x^2 – x + 3


Now let us assume some values of x. Therefore let x be equal to -2, -1, 0, 1, 2, 3, 4 one by one. Now we shall find the value of y for all these values of x by plugging the x values into the above equation. So, for x = -2,

Y = (-2)^3 – 3(-2)^2 – (-2) + 3 = -15. So our first ordered pair is (-2,15).

Next for x = -1,

Y = (-1)^3 – 3(-1)^2 – (-1) + 3 = 0. So our second ordered pair is (-1,0)

Similarly the ordered pairs for the remaining values of x would be: (0,3), (1,0), (2,-3), (3,0), (4,15)



Now we shall plot these points on a graph sheet and run a smooth curve through the points. That would be our required graph.