The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

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/ 
Associated Topics:
High School Equations, Graphs, Translations
High School Functions

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/