Question

Given `f(x) = -7x^2(2x-3)(x^2+1)` how do you determine the following?

a) The degree of `f(x)`.
b) The leading coefficient.
c) The maximum possible number of turning points.
d) The real zeros of `f(x)`.
e) The end behaviour of `f(x)` as `x->-oo`
f) The end behaviour of `f(x)` as `x->oo`

Answer 1

a) `" "5`
b) `" "-14`
c) `" "4`
d) `" "0` of multiplicity `2`, `3/2` of multiplicity `1`
e) `" "f(x)->oo` as `x->-oo`
f) `" "f(x)->-oo` as `x->oo`

Explanation:

Given:

`f(x) = -7x^2(2x-3)(x^2+1)`

It is not too arduous to multiply out `f(x)` fully, but we do not need to in order to answer all of the questions.

In particular, note that the term of highest degree in `f(x)` comes from multiplying:

`-7x^2(2x)(x^2) = -14x^5`

a) The degree of `f` is the highest degree of any term, i.e. the degree of `-14x^5`, which is `5`. It is a quintic.

b) The leading coefficient is the coefficient of this term, namely `-14`.

c) A polynomial of degree `n > 0` has at most `n - 1` turning points. So our quintic has at most `4` turning points.

d) The real zeros of a polynomial correspond to its linear factors. In our example `f(x)` has linear factors `x`, `x` and `(2x-3)`, corresponding to real zeros `0`, `0` and `3/2`. The quadratic factor `(x^2+1)` will be non-zero for any real value of `x`, since `x^2 >= 0` for any real value of `x`. The coincident zeros for `x=0` are described by saying that `f(x)` has a zero at `x=0` of multiplicity `2`. The zero at `x=3/2` has multiplicity `1`.

e) As `x->-oo`, the behaviour of `f` is determined by the leading term `-14x^5`, which becomes much larger faster than the other terms. Since this term is of odd degree and the coefficient is negative (`-14 < 0`), `f(x)->oo` as `x->-oo`.

f) As `x->oo`, the behaviour of `f` is dominated by the term of highest degree `-14x^5`, which becomes large and negative. So `f(x)->-oo` as `x->oo`.

graph{-7x^2(2x-3)(x^2+1) [-5, 5, -25, 25]}