# given the polynomial f(x)=-x^2(x+4)^3(x^2-1) find the following the zeros & multiplicity of each where the graph crosses / touches the x axis number of turning points end behavior

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### 1 Answer

Given the function `f(x)=-x^2(x+4)^3(x^2-1)`

we find the zeros and their multiplicities by looking at the factors of the function. Setting each factor to zero gives the zeros, and the multiplicity is from the exponent of the factor.

This means that one zero is `x=0` with multiplicity 2 from `x^2` .

Another is `x=-4` with multiplicity 3 from `(x+4)^3` .

The last zeros are `x=+-1` with multiplicity each of 1 from `x^2-1=(x-1)(x+1)` .

This means that at x=0, the graph touchs the axis, and at x=-4, 1, -1, the graph goes through the axis.

**Turning points:**

The turning points are found from the second derivative which has degree 2 less for polynomials. In this case, the polynomial will go from degree (2+3+2)=7 to having second degree 5, which means there will be 5 turning points.

**End Behaviour.**

The end behaviour is found from the coefficient of the leading terms. Since the polynomial is degree 7, it behaves like `-x^7` at infinity. This means that as `x->infty` , `y->-infty` and as `x->-infty` , `y->infty` .