SOLUTION: How to find the end behavior of this function? x(2-x)(x+3)^2 would it be: as x approaches positive or negative infinity, y approaches negative infinity?

Question 583247: How to find the end behavior of this function?
x(2-x)(x+3)^2
would it be: as x approaches positive or negative infinity, y approaches negative infinity?
Found 2 solutions by solver91311, lwsshak3:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

First of all, if you are going to say that it is a function, write it like a function, namely:

Then re-write to expand the squared binomial:

Now, evaluate each of the factors for its sign if is a relatively large positive number, say 100.

First factor is +
Second factor is -
Third factor is +
Fourth factor is +

Plus times Minus times Plus times Plus is Minus. As increases without bound, i.e. , decreases without bound, i.e. .

Repeat the process with a relatively large negative number, -100 should do nicely:

First factor is -
Second factor is +
Third factor is -
Fourth factor is -

Minus times Plus times Minus times Minus is Minus. As decreases without bound, i.e. , decreases without bound, i.e. .

John

My calculator said it, I believe it, that settles it

Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
How to find the end behavior of this function?
x(2-x)(x+3)^2
**
zeros are: 0, 2 and -3 (multi 2)
Draw a number line with the zeros on it to show how the sign of f(x) changes as we go thru the zeros.
<..-...-3....-.....0.....+....2....-....>
When x is >2, f(x)<0
Moving to the left, signs of the intervals switch as we go thru zeros of odd multiplicity like 1, 3, 5 etc. but do not switch when going thru zeros of even multiplicity like 2, 4, 6 etc. For this problem, 3 is of even multiplicity 2, while 0 and 2 are of odd multiplicity 1.
You are right that as x approaches negative or positive infinity, f(x) approaches negative infinity.
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