Find the zeros of the polynomial function and state the multiplicity of each

[pita0001]

f(x)= x (x+2)^2 (x-1)^4Zeros would be: 0, -2, 1
Multipicity of : 1 2 4Then for y- intercept: f(0)=0

And don't know how to graph it...

 

[MarkFL]

What is the difference in the way a polynomial behaves at a root of even multiplicity as opposed to a root of odd multiplicity?

 

[pita0001]

Do you mean like when graphing them?

Because I know odd goes through points and even only touches.

 

[MarkFL]

Right, at a root of even multiplicity a polynomial touches the $x$-axis but does not pass through it. So, that is one thing we can use to sketch the graph. What is the degree of the given polynomial, and what is the difference between polynomials of even and odd degrees with respect to their behavior as \(\displaystyle x\to\pm\infty\)?

 

[pita0001]

Would my graph be like:

going from positive infinity to -2, then down making a U and going through zero then down and going through 1, ending going up towards positive infinity?

(Don't know how to draw it on here)

 

[MarkFL]

The given polynomial is 7th degree, and polynomials of odd degree behave as follows:

\(\displaystyle \lim_{x\to-\infty}f(x)=-\infty\)

\(\displaystyle \lim_{x\to\infty}f(x)=\infty\)

So, you know on the far left the function is going to negative infinity and on the far right it is going to positive infinity.

That means it approaches the root at $x=-2$ from below, just touches the $x$-axis then goes back down before coming back up to pass through the origin, continues up, then comes back down to touch the $x$-axis at $x=1$ then shoots off unbounded towards positive infinity.

To find the exact locations of the turning points would require the calculus. But, you have enough information now to make a reasonable sketch of the graph. :D

 

[HallsofIvy]

Another way to think about this is to look at signs in each interval.

You have $$f(x)= x (x+2)^2 (x-1)^4= (x- 0)(x- (-2))^2(x- 1)^4$$.

I have written each as "x- " because "x- a" is negative if x< a, positive if x< a.

Here the terms are 0 at x= -2, 0, and 1. If x is any number less than -2, then it is less than all three so each of x, x+ 2, and x- 1 is negative. The even power, of course, is positive anyway so we have (-)(+)(+)= - for x< -2. If -2< x< 0, the x+ 2 term is now positive but since it was to an even power it was positive any way so we still have (-)(+)(+)= -. The graph goes up, touches the x-axis at (-2, 0) then goes back down again. It must turn some where (do you know how find minimum points? That may require calculus) to come back up to 0. For 0< x< 1, the x term is now positive so we have (+)(+)(+)= +. The graph continues up until it turns back to go to 0 at x= 1. For 1< x, all three terms are still positive: (+)(+)(+)= + so the graph must turn back up.

 

[Deveno]

In slightly less technical terms, the curve start very negative from the left, and grows until it has a "hump" tangent to the $x$-axis at $x = -2$.

It has a valley in-between -2 and 0, and another "hump" or "peak" between 0 and 1. It then has another valley which "bottoms out" at the $x$-axis at $x = 1$, after which it grows forever more.

With a 7-th degree polynomial, it is possible to have 3 "peaks" and 3 "valleys", but in point of fact, that does not happen here (as the derivative only has 2 roots the original function does not, and 0 is not a root of the derivative, giving only 4 critical points).

The valley from -2 to 0 is considerably deeper than the peak from 0 to 1 is tall. It turns out that the bottom of that first valley is around -18, while the top of the second peak is only around 0.4. So on the entire interval [0,1], the curve is "almost flat", staying within 1/2 of the $x$-axis.

 

[soroban]

Hello, pita0001!

Your description is inaccurate.

Would my graph be like:
Going from positive infinity to -2, then down making a U
and going through zero, then down and going through 1,
ending going up towards positive infinity?

The graph starts at the far left, down at negative infinity.
It rises and is tangent to the x-axis at $$x = -2.$$
It turns down, then turns up and passes through the origin.
Then it turns down and is tangent to the x-axis at $$x = 1.$$
Then it rises, soaring off to positive infinity.

                  |
                  |         *
                  |
                  |  *     *
         -2       |*  *   *
    - - - * - - - * - - * - -
        *   *     |     1
       *     *   *|
               *  |
      *           |
                  |