Writing a Polynomial with similar end behavior, using only two coordinates: A zero and the y-intercept.

JayChampion

JayChampion

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Recently, my class began a unit, in which we are primarily working in functions, and one of the segments is all about writing polynomials with similar end behaviors based on the previous function given, and even though we went through the segment, I even asked my teacher for help, but I still don't get it...
The one of the problems from my latest worksheet is as follows:
b) Write a polynomial g(x) with the same end behavior as f(x) that crosses the x-axis at x=2 and bounces at x=-3
The equation f(x) is: x^5-10x^4-9x^3+28x^2+20x-2
 
I forgot to mention, since there CAN be multiple answers, there could be multiple ways to solve, but I couldn't find anything, trying quadratic formula, to attempt to simplify to then get an answer that I (might) be able to move over to g(x), but that failed.
 
First, you have to identify the end behavior of the given f(x). Can you describe that (in whatever terms are standard in your class)?

Then you have to write a polynomial with that end behavior, with x-intercept x=2 that crosses, and another at x=-3 with the multiplicity that will cause the graph to "bounce".

What does the end behavior tell you about the degree of your g(x)?

What do the intercepts and their multiplicity (bouncing) tell you about one factor of g(x)?

Once you answer those questions, you can fill in any extra factors you might need any way you like. That's where multiple answers can come in. But there's one simplest answer that doesn't require much extra work, if any.

Tell me whatever you can in answer to my questions, and we can get moving toward an answer.
 
Dr Peterson:
End behavior would be:
x>-(infinity), f(x)>-(infinity)
x>(infinity), f(x)> (infinity)
The end behavior tells me that it should be odd, since it ends going the opposite directions.
The intercept tells me that the one factor will be (x+3)^2

Romsek:
Whoops, I wrote it wrong, it should say "y-axis", instead of "x-axis"
It should cross the y-axis at (0,2).
Sorry if it caused you to waste some of your time.
 
Also, I do know that there will be three other factors that will be use here, but what they are, I don't know. I wouldn't know how to find them.
 
JayChampion said:
Dr Peterson:
End behavior would be:
x>-(infinity), f(x)>-(infinity)
x>(infinity), f(x)> (infinity)
The end behavior tells me that it should be odd, since it ends going the opposite directions.
The intercept tells me that the one factor will be (x+3)^2

Romsek:
Whoops, I wrote it wrong, it should say "y-axis", instead of "x-axis"
It should cross the y-axis at (0,2).
Sorry if it caused you to waste some of your time.
Click to expand...
Are you saying that where you said "crosses the x-axis at x=2", you meant "crosses the y-axis at y=2"?

I'm going to assume that what you wrote originally in this thread is correct, as you seem to be responding to something Romsek said elsewhere about a different problem.

You want an odd degree, so let's make it 3 to keep things simple.

You have one squared factor, (x+3)^2. You only need one more factor (and possibly a constant multiplier).

If it has a crossing x-intercept at x=2, that gives you your third factor. Put it together, check your answer against the requirements, and you're done.

If you are not given another x-intercept, but a y-intercept, then pick an x-intercept at random, then find a constant multiplier that will give the required y-intercept.
 
I had a thought about trying quadratic formula, but it would help me find the factors, if I had an equation, but I do not have the equation.
 
JayChampion said:
I had a thought about trying quadratic formula, but it would help me find the factors, if I had an equation, but I do not have the equation.
Click to expand...

How would you use the quadratic formula??

Remember, you are making up a function g, and beyond the intercepts you are given, any others are up to you to choose -- you don't have to find any intercepts. This is a creative exercise, up to a point, in which you have a certain amount of freedom. That lets you relax.
 
Yes, sorry, that was to another board. Ok, that actually sounds pretty simple when you put it that way, I'm going to try to see if that works.
 
I don't know, I was simply trying to find a way to get the answer. I guess I had my mind stuck on getting a fifth power. Since what you said was correct, it was truly far simpler than I was making it.
 
Thank you, it might have just been the problem, or it might be me, but, again, thank you for helping me understand it!
 
If you did want to make a fifth degree polynomial, you could choose two more intercepts arbitrarily, so although it would be more complicated, you still wouldn't have to worry about what to choose! I imagine you are not accustomed to math problems in which you are allowed to make arbitrary choices (such as what degree to use). Teachers can forget that students haven't seen this kind of problem before, and fail to explain that fully!
 
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