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#

1 Answer
Aug 12, 2017

Answer:

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]}