Shapes and Polynomials: Artful Curves

An appendium to section 3.2 of the MS 100 College Algebra text book: Polynomial functions of higher degree and their graphs, Algebra and Trigonometry by Larson, Hostetler, sixth edition.

Polynomials can be designed to produce specific shapes. The following is a shape that my son refers to as "batman."

Generating a particular shape requires knowing something about polynomials behavior. The first is that the above "opens down" with the "end behavior" being that the graph "enters and leaves" from the bottom of the graph. This means we need an even function such as f(x) = x².

Even functions are functions in which the largest exponent is an even number. The simplest even function is f(x) = x². f(x) = x² enters and leaves from the top of the graph as seen below on the left, but f(x) = −x² enters and leaves from the bottom, as seen below on the right. This is due to the order of operations: the square is applied to the x first, then the negative sign is applied to the squared value. This is a beginning for building the "bat" function above.

Note the minus signs were cut off from the y-axis on the graph at the above right. Note also that the "lead coefficient" for f(x) = −x² is actually (−1) as in f(x) = (−1)(x² ). Page 273 notes this impact of the sign of the lead coefficient.

To get a graph to pass through x = ± 5, the graph must have roots (x-intercepts) at x = ± 5. Remember that for the purposes of this course that roots, x-intercepts, zero's of a function, and solutions of a function all mean the same thing: the place a graph either crosses or touches the horizontal x-axis. This is also explained on page 275 of the text.

To get an x-intercept at x = −5, then a factor of (x + 5) is needed. This is because x = −5 can have 5 added to both sides to produce x + 5 = 0, which leads back to the factor (x + 5) = 0. Likewise, x = 5 has corresponding factor (x − 5) = 0. Thus f(x) = −(x + 5)(x − 5) or f(x) = −x² + 25 yields x-intercepts of x = ± 5. This can be seen in the graph on the left below.

To get a graph to touch the x-axis and the "double-back," to "touch and return," then one needs a double set of x-intercepts at the same location. This can be seen in the graph on the above right. This is the result of the following function:

f(x) = −(x + 5)(x + 5)(x − 5)(x − 5)

Note the repetition of the factors – this causes the "touch and return" behavior at x = ± 5. The text book calls these "touch and return" points "repeated factors" on page 275. Note too the lead negative to force the function to open down. The above function can be written two other ways:

f(x) = −(x + 5)²(x − 5)² or as f(x) = −(x² − 25)(x² − 25)

Try that at home, see if you can convert from −(x + 5)(x + 5)(x − 5)(x − 5) to the two forms above. A third way to write this same double touch at x = ± 5 is multiply the factors out completely:

f(x) = −x⁴ + 50x² − 625

Only in this fully "expanded" form can we determine the y-intercept which is y = −625 as seen on the graph. Can you multiply the original factors out to get this result?

To get the crossing of the function at x = ± 2 we need two more factors:

f(x) = −(x + 5)(x + 5)(x − 5)(x − 5)(x + 2)(x − 2)

This can be expanded to form:

f(x) = −x⁶ + 54x⁴ − 825x² + 2500

This now looks like the graph on the right.

The final touch is to get the very middle of the graph, the present axis of symmetry, to do a touch and return on the origin at (0, 0). At this point graphing really requires a computer as the numbers are getting too large to work by hand. Note that the curves are smooth curves, something that would not be revealed if one plotted only the integer values of the function.

To get a touch and return at the origin the function must have a double root at x = 0, or two factors that are both (x − 0 ). Note that (x − 0 )(x − 0 ) equals x² − 0x + 0 or just plain old x².

Thus the function for the original bat curve is:

f(x) = −(x + 5)(x + 5)(x − 5)(x − 5)(x + 2)(x − 2)(x − 0 )(x − 0 )

or

f(x) = −(x + 5)(x + 5)(x − 5)(x − 5)(x + 2)(x − 2) x²

more often written:

f(x) = −x²(x + 2)(x − 2) (x + 5)² (x − 5)²

This multiplies out to:

f(x) = −x⁸ + 54x⁶ − 825x⁴ +2500x²

This function generates the original "bat" function seen at the start of this paper.

Some parting ideas:

• To open down one needs an even function with a negative lead coefficient

• To open up one needs an even function with a positive lead coefficient

• To cross the x-axis one needs a factor corresponding to the crossing value

• To touch and return from the x-axis one needs a factor squared that corresponds to the value at the touch and return point.

• The use of x-intercepts that are equally distant left and right from the y-axis generates symmetric polynomials with the y-axis as the axis of symmetry such as those seen in this paper.

Produced using GnuPlot 4.0, Maxima 5.9.2, and OpenOffice.org 2.0.2.

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- Dana Lee Ling, spring 2006