Curvy ... and Topsy-Turvy?
Date: 11/22/2014 at 23:06:17 From: Sam Subject: Understanding the Behaviour of Curves When sketching curves of higher degree polynomials, we can use factorisation to look at the parts of the equation that make up the whole polynomial. Therefore, we can look at the curve in terms of smaller curves, especially on the x-axis (the roots). For example, consider the function y = (x - 1) * (x + 3)^2 * (x - 2)^3 * (x + 1)^4 Its curve will cross the x-axis like a linear function at 1, like a parabola at -3, like a cubic at 2, and like a quartic at -1. When we do this, we get the shape of the curve. It begins in the first quadrant and ends in the second because it is positive. But when we look at the roots of the equation, the linear function formed at (1, 0) looks like the equation y = -x - 1 (negative) rather than y = x - 1 (as in the original term). Why is this? Even though the original term was (x - 1), the graph of the entire function looks negative near x = 1. Similarly, in other regions, the signs of the roots seem opposite to the roots of the individual terms. What part of the equation determines this?
Date: 11/22/2014 at 23:38:33 From: Doctor Peterson Subject: Re: Understanding the Behaviour of Curves Hi, Sam. An excellent question! Let's look at the factor (x - 1) in its context: y = (x - 1) * (x + 3)^2 * (x - 2)^3 * (x + 1)^4 ===== Since we're looking at the curve NEAR x = 1, all the other factors will be near the value they take at x = 1. (A small change in x will produce a large relative change in (x - 1), since it is near zero, but only a small relative change in the other factors.) If we replace x with 1 in those other factors, we have y =~ (x - 1) * (1 + 3)^2 * (1 - 2)^3 * (1 + 1)^4 = (x - 1)(16)(-1)(16) = -256(x - 1) That factor of -256 answers your question. The specific linear function that approximates the curve near x = 1 is y = -256(x - 1), with a negative slope. Try doing the same with the other x-intercepts, and graph these approximations on the same axes as the curve. I think you'll like what you see! Here is my graph: - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 11/23/2014 at 00:02:08 From: Doctor Greenie Subject: Re: Understanding the Behaviour of Curves Hi, Sam -- What you are saying is not true at all. The factorization of the polynomial tells us where the roots are, and the powers of the factors tell us the general behavior of the function at the zeros. But it appears that you are saying that the function in your example, because of the factor (x - 1), should look like the linear function y = x - 1 at x = 1. That is not at all true. If it were, then the slope of the function at every zero would be positive, because the factors are all of the form (x - a), not (-x - a), and that would not be possible. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
Date: 11/23/2014 at 14:49:08 From: Doctor Peterson Subject: Re: Understanding the Behaviour of Curves [fixed] Hi, Sam. I have a clarification to make. I tell my students that a polynomial near any zero h "looks like" the corresponding factor (x - h)^n, in the broad sense that if the factor is cubed, it looks like a cubic, and so on. This is what you said initially ("like a cubic"). A cubic, of course, can look either like x^3 or like -x^3, which is where your question arose. In a more exact sense, it looks like a(x - h)^n, where the coefficient "a" has to be determined as I showed you, by putting x = h into the other factors. If you want to get the direction right, as well as the overall shape, you have to at least determine the sign of this coefficient. So the answer to your specific question -- about what part of the expression determines the direction -- is: "all the rest." The graph I shared yesterday showed this for all four zeros, carrying out the plan I suggested. Note how, with the coefficients calculated correctly, the graph looked very much like the corresponding functions; in fact, no polynomial could more closely "snuggle up to" the curve at those points than these. They are called "osculating curves" (from the Latin for "kiss"): http://mathworld.wolfram.com/OsculatingCurves.html - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.