polynomial functions

Question

Find a factored form for the polynomial function f that has minimal degree. Assume that the intercept values are integers and that Xscl = Yscl = 1.

 

Normally, I could solve this problem, but the graph I am giving is from a graphing calculator, and I can't read it. I need to find the zeroes before I can find it, yet can't read or understand the graphing calculator's graph. I think that it bounces off the x axis at -1 and 1, and it is sloping down, so it starts and ends below the x axis. It starts coming up from below, bounces off x=-1, goes back down and turns around at y=-2, and then goes back up, and bounces off x=1, and then goes back down.

 

Please help. Thanks!

Answer 1

Hi James,

 

There are a few concepts here that you want to make sure that you understand so that you can answer this and future problems of this sort.  

 

First - End Behavior. Because you say the graph begins from the bottom and end at the bottom, then this tells us the function is an EVEN function because both ends face the same direction. [An Odd on the other hand means one is at the top and the other is at the bottom.] Now, to determine if it is a Positive or a Negative Function we look at the direction the graph finishes, if going upwards its considered a positive graph, in this case, because it finishes downwards we know this i a NEGATIVE graph so we will have a "-1" multiplied at the front of our equation.

 

Zeroes - Each time the graph hits on the x-axis that tells us about a zero and thus one of the factors. You state the graph hits at x = -1, so that means we have a "zero" at -1 and the factor (x+1) will be a part of our function. Same thing goes for (X-1) because of the other "zero" we were given.

 

Multiplicity - When we look at the "zeroes" on a graph there are 3 types: Where the graph Crosses, Bounces and Squiggles through those x-axis values and each one tells us something special, this is called "multiplicity".  If the graph crossed the x-axis at x=5 for example we would say (x-5) is the factor and the exponent on it is 1. If the graph bounced off instead we would say (x-5)² where the exponent is 2, and then lastly if it crosses not as a straight line but "squiggles" then we say that would have a muliplicity of "3" so we would see (x-5)³ instead. 

For our problem you said the graph bounces at x=-1 (multiplicity of 2) and bounces at x=1 (also multiplicity of 2) this gives us: (x+1)²(x-1)²

 

Put it all together - Now that we have the correctly multiplicity for the given zeroes now we get to put that together with the end behavior that was originally discussed, because BOTH (even power graph) point downward (negative function) we said the function would have a "-1" multiplied in front.

 

Final Equation: f(x) = -1(x+1)²(x-1)²

We can identify the zeros, we can tell both zeros bounce and we can see that the result is a downward, even power graph. 

 

Hope this helps. It may have more words than you expected but you should be able to us this for any future Polynomial problem as well. ;)