# find the zeros and describe the end behavior of f(x)=2x(x-1)(x+1). is f(x) odd, even, or neither? explain.

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Find the end behavior of y=2x(x-1)(x+1):

This is a cubic polynomial with positive leading coefficient. (You can write in standard form: y=2x(x+1)(x-1) ==> `y=2x(x^2-1)`

==>`y=2x^3-2x` The leading coefficient is 2.

A cubic with positive leading coefficient has the following end behavior:

As `x->-oo,y->-oo` and as `x->oo,y->oo` .

(As x decreases without bound, y decreases without bound; as x increases without bound, y increases without bound.)

The graph:

(Note that the function is odd -- there is rotational symmetry of 180 degrees about the origin.)

f(x)=2x(x-1)(x+1)

The zeros of f(x) is 0,1 and -1.

x=0 ,

x-1=0

x=1

and

x+1=0

x=-1

Thus function has three zeros.

`f(x)=2x(x-1)(x+1)`

`=2x(x^2-1)`

`=2x^3-2x`

`f(-x)=-2x^3+2x=-(2x^3-2x)=-f(x)`

`f(-x)=-f(x)`

Thus by def of odd function. f(x) is an odd function.