Expert: Scotto - 7/26/2008

Question
how to find intercepts on graphs using cubics?  

Answer
There is no simple way to do this.

Method 1: Take the derivative, find the maximums, and use the bisection method to find the roots between the maximum and minimum (if max is positive and min is negative).

There are also two values to be found just outside the maximum and minimum.  Take the x value at the minimum and an x value far enough way to be positive.  Use the bisection method to find the root.  Do the same with the maximum.

Using these, the intercept in between the maximum and minimum can be found (if there is one) and the intercepts on the outside (above the maximum and below the minimum) can be found.  Note that if the coefficient on x^3 is negative, the roots will be found below the maximum and above the minimum.


Method 2: Taken out of mathforum.org/dr.math/faq/faq.cubic.equations.html.
Note that this method might be able to find a more exact solution, but is much more complicated.

Let's say you have the equation ax3 + bx2 + cx + d = 0. The first thing to do is to get rid of the "a" out in front by dividing the whole equation by it. Then we get something in the form of x3 + ex2 + fx + g = 0. The next thing we do is to get rid of the x2 term by replacing x with (y - e/3). The new equation y3 + py + q = 0 has no y2 term. This is much easier to solve, although it's still kind of hard and it took a lot of really smart people a long time to do.

Let's introduce two new variables, t and u, defined by the equations

            u - t = q               and   
            tu = (p/3)3.

If you're really interested, you might want to know what t and u actually are in terms of p and q:

        -3qSqrt{3} +- Sqrt{4p3 + 27 q2}
    t = ---------------------------------
                    6 Sqrt{3}
          
         3qSqrt{3} +- Sqrt{4p3 + 27q2}
    u =  --------------------------------
                    6 Sqrt{3}

That +- is either + or - for BOTH of the equations at the same time. You can't have one as a + and the other as a - .

Anyway, y1 = CubeRoot{t} - CubeRoot{u} will be a solution of y3 + py + q = 0.

Verify this result now (DON'T use the explicit formulas for t and u, just plug CubeRoot{t} - CubeRoot{u} into the cubic polynomial), and make sure you see why it works. To find the other two solutions of our cubic polynomial (if there are any real ones) we could divide y3 + py + q by its known factor (y - CubeRoot{t} + CubeRoot{u}), getting a quadratic equation that we could solve by the quadratic formula. Finally, once we have the solution values of y, we can find x = y - e/3.

As you can see, the complexity is much greater than in the Quadratic Formula. To try to go backward and come up with a closed form for the Cubic Formula in terms of the original a, b, c, d would be a real pain.

For a nice reference about this problem, as well as some information about the Quartic, look at the 3-volume set called Mathematical Thought from Ancient to Modern Times by Morris Kline. It gives the math and the history together.

For more about solving cubic polynomial equations, see Another Solution of the Cubic Equation, which describes an approach (submitted independently by Paul A. Torres and Robert A. Warren) based on "completing the cube."