# Behavior of polynomial functions at their zeros

• B
Gold Member
Just a general question here.
So for a polynomial function, the behavior of the graph at the zeros is determined by the evenness or oddness of the magnitude of the zeros. If the magnitude is odd, the graph will cross the zero. If the magnitude is even, it will bounce at the zero. Why is this true? What specifically causes these different actions?

Also, I'm trying to figure out why even degree functions look like a parabola, but odd degree functions look like an S shape? I understand what causes the parabola, but I'm not sure what causes the S shape. What causes it to steadily climb, then decrease the rate at which it climbs, then goes back to climbing again at the original rate?

Mark44
Mentor
So for a polynomial function, the behavior of the graph at the zeros is determined by the evenness or oddness of the magnitude of the zeros. If the magnitude is odd, the graph will cross the zero. If the magnitude is even, it will bounce at the zero. Why is this true? What specifically causes these different actions?
A better choice of word than "magnitude" is "exponent."
Take a look at the graphs of ##y = x^2##, ##y = x^4##, and other power functions with an even exponent. Although only the first one I listed is a parabola, all have a similar shape, with the difference being for the exponents of 4 and higher, the graphs are flatter between -1 and +1, and steeper for values less than -1 or greater than +1. If x = a is a zero of a polynomial, then there will be a factor of x - a to some power in the factored form of the polynomial. In a small region around a, the (x - a)n factor will play a dominant role in the behavior of the graph. Outside this region, the other factors play more dominant roles.

Also, I'm trying to figure out why even degree functions look like a parabola, but odd degree functions look like an S shape? I understand what causes the parabola, but I'm not sure what causes the S shape. What causes it to steadily climb, then decrease the rate at which it climbs, then goes back to climbing again at the original rate?
Take a look at ##y = x##, ##y = x^3##, and other power functions with an odd exponent. Since they are all to an odd power, if x < 0, then y < 0, and if x > 0, then y > 0. With y = x you get only a straight line through the origin, but with ##y = x^n## where n is an odd integer, you get the S-shape. Between -1 and 1 you have a number less than 1 in magnitude, and when you cube or raise this number to a higher odd power, the result will be even smaller in magnitude. For example ##(\frac 1 2)^2 = \frac 1 4 < \frac 1 2##. Same for all numbers between -1 and 1.

• opus
Gold Member
Shoot. I meant to say "multiplicity" not magnitude. That is, ##\left(x-2\right)^2## = 2 with a multiplicity of two.
So ##x^4## is not considered a parabola? I thought that it was. I can see what you mean by flattening out. To my understanding, this is because as ##\left|x\right|## gets higher and higher, the x terms get closer to zero.
For the zero at x - a, the graph will either cross or bounce at this value. If the multiplicity of x- a is even, it will bounce, and it it's odd it will cross. Why is this so?

And I completely understand now with the explanation on S shaped graphs. That makes a lot of sense! Very cool!

Mark44
Mentor
Shoot. I meant to say "multiplicity" not magnitude. That is, ##\left(x-2\right)^2## = 2 with a multiplicity of two.
So ##x^4## is not considered a parabola?
Right, it's not a parabola.
opus said:
I thought that it was. I can see what you mean by flattening out. To my understanding, this is because as ##\left|x\right|## gets higher and higher, the x terms get closer to zero.
Not sure what you're trying to say here. If ##|x| < 1##, then ##|x|^2 = x^2 < |x|##, and the same is true for ##x^4, x^6##, etc. This is the reason for the flattening out between -1 and +1.
opus said:
For the zero at x - a, the graph will either cross or bounce at this value. If the multiplicity of x- a is even, it will bounce, and it it's odd it will cross. Why is this so?
For the same reason that ##y = x##, ##y = x^3##, and other odd-exponent power functions cross the x-axis at 0, and that ##y = x^2##, ##y = x^4##, and other even-exponent power functions just touch the x-axis at 0.

Try out an example to see this. Let ##y = (x - 1)(x - 2)^3(x - 5)^2##
Look at the behavior near x = 2. Using a calculator, what y values do you get for these x-values?
x = 1.9
x = 1.95
x = 1.98
x = 2.02
x = 2.05
x = 2.1

Of course all the y-values are going to be near zero, but do these values cross through the y-axis or bounce off it?

Try something similar near the root x = 5.

opus said:
And I completely understand now with the explanation on S shaped graphs. That makes a lot of sense! Very cool!

Gold Member
Right, it's not a parabola.
Not sure what you're trying to say here. If ##|x| < 1##, then ##|x|^2 = x^2 < |x|##, and the same is true for ##x^4, x^6##, etc. This is the reason for the flattening out between -1 and +1.
For the same reason that ##y = x##, ##y = x^3##, and other odd-exponent power functions cross the x-axis at 0, and that ##y = x^2##, ##y = x^4##, and other even-exponent power functions just touch the x-axis at 0.

Try out an example to see this. Let ##y = (x - 1)(x - 2)^3(x - 5)^2##
Look at the behavior near x = 2. Using a calculator, what y values do you get for these x-values?
x = 1.9
x = 1.95
x = 1.98
x = 2.02
x = 2.05
x = 2.1

Of course all the y-values are going to be near zero, but do these values cross through the y-axis or bounce off it?

Try something similar near the root x = 5.
I may be thinking of something else, but what I was referring to is:
If we have a polynomial, say ##P\left(x\right) = 2x^3+5x^2-7x+11##, by the distributive property, ##x^3\left(2+ \frac 5 x - \frac {7}{x^2} + \frac {11}{x^3}\right)##. As ##\left|x\right|## gets very large, the terms inside the parentheses get close to zero and ##P\left(x\right)≈2x^3## It was to my understanding that this is the reason the graph flattens out.

I don't think I'm entirely sure on what you're saying with the x values. I graphed the function and can see that y values very close to, but to the left of x=2 are negative, and once the graph passes x=2, the y values become positive.

Mark44
Mentor
Since the thread title is "behavior of the graphs of polynomials near their zeros, that's the context in which I was responding.
I may be thinking of something else, but what I was referring to is:
If we have a polynomial, say ##P\left(x\right) = 2x^3+5x^2-7x+11##, by the distributive property, ##x^3\left(2+ \frac 5 x - \frac {7}{x^2} + \frac {11}{x^3}\right)##. As ##\left|x\right|## gets very large, the terms inside the parentheses get close to zero
All the terms inside get close to 2, not zero.
opus said:
and ##P\left(x\right)≈2x^3## It was to my understanding that this is the reason the graph flattens out.
For large |x|, the graph doesn't flatten out -- it gets close to the graph of y = 2x^3. By "flattening out" I meant that there were stretches where the graph is nearly horizontal.
We're really talking about two different things. I was talking about the behavior of the graph near one of the roots, and you're talking about long-term behavior -- what the graph starts to look like for large |x|.
opus said:
I don't think I'm entirely sure on what you're saying with the x values. I graphed the function and can see that y values very close to, but to the left of x=2 are negative, and once the graph passes x=2, the y values become positive.
What that is saying is that "near x = 2" the graph behaves like the graph of y = K(x - 2)^3; that is, a cubic shifted to the right by 2 units. The value of K is determined by the other factors in the polynomial. As far as K is concerned, all you really need to get is the sign of K. If it's positive, the graph looks like y = (x - 2)^3. If K is negative, the graph looks like y = -(x - 2)^3. Once you get away from each root, the other factors become more dominant.

Using this line of thinking, you can get a good idea of what the graph of your polynomial looks like near each root, and then to get the full graph, just connect these curves.

mathwonk
Homework Helper
2020 Award
Take a typical example like Y = (X-1)(X-2)(X-3)(X-4)(X-5). First of all note that the graph has height Y = 0, i.e. it crosses the X axis, exactly when X = 1,2,3,4 and 5. I.e. the “roots” are exactly 1,2,3,4 and 5. So we just have to understand the behavior in between those points. The simplest thing would be that the graph crosses the X axis at each of these points. Hence, since there is an odd number of these points, it must start out on one side of the X axis and end up on the opposite side. Since Y is positive for very positive X, it must end up above the X axis and hence must start out below it.

Now, to see that this is indeed what happens, look at the value of each factor at a given value of X, and let X move along the X axis from left to right, and ask what is the value of Y.

Now if X is less than a, then the factor (X-a) is negative, while if X is greater than a then (X-a) is positive. So as X moves along the X axis from left to right, each of the 5 factors gradually changes from negative to positive. At each stage, the sign of Y depends on the number of negative factors (X-a), i.e. on how many of the roots is larger than the given value of X.

I.e. for X less than 1, all 5 factors (X-1), (X-2), (X-3), (X-4), (X-5), are negative, so their product is also negative, hence Y is negative everywhere to the left of X=1. Then as X moves to the right, and passes slightly to the right of 1, the factor (X-1) becomes positive, but all 4 of the other factors are still negative. So for X between 1 and 2, there is one positive factor and 4 negative factors, so the product Y is positive and the graph is above the X axis.

In this way we can see that each time X passes just to the right of another root, the number of negative factors becomes one less and the sign of Y changes, so the graph crosses the X axis. Hence with an odd number of factors, the graph styartts out on one side of the X axis and ends up on the opposite side.

Now let the roots move as well, say let the first root X=1 get larger, becoming 1+e, as e grows from 0 to 1, i.e. let the first root grow from 1 to 2, so that the first two roots become equal. As X grows closer and closer to 2, the amount by which the graph rises above the X axis between the first two roots shrinks, so the the little bump between the first two roots shrinks down the meet the X axis when the first two roots become equal.

When the first two roots become equal and our equation becomes

Y = (X-2)(X-2)(X-3)(X-4)(X-5) = (X-2)^2(X-3)(X-4)(X-5), we can see that now, that if we again let X grow from left to right, now Y is negative for X less than 2, but as X crosses from below 2 to above 2, both of the factors (X-2) change sign. Then there are two positive and three negative factors, so the product is still negative, forcing the graph to “bounce” when it hits the X axis and go back below it, until X reaches 3. Thus the graph bounces at X=2 but crosses the X axis at every later root, again ending up above the X axis after X passes above 5.

Any combination of combined roots for a polynomial of degree 5 (and led coefficient = 1), can be visualized by pulling on this graph like a string so that various roots come together.

If we keep pulling down on the graph after the first two roots come together, we can pull the first bump of the graph down below the X axis entirely. Then the product of the first two factors goes from (X-2)^2 to (X-2)^2 + e^2 = (X-2-ei)(X-2+ei). I.e. the first two roots go from the real pair 2,2 to the complex pair 2+ei, 2-ei. These two roots do not appear on the real graph, and we lose one “bump” from the real graph.

We can do this once more, making another pair of real roots complex, but we then must still have one real root left. The graph of the 5th degree polynomial thus can cross the X axis only once. In fact we can even eliminate all 4 bumps from the graph by pulling on it, until the graph just goes consistently up, like that of Y = X^5 + 1 = (X+1)(X^4-X^3+X^2-X+1).