How do you describe the end behavior of #y=8-x^3-2x^4#?

2 Answers
May 28, 2018

Answer:

See explanation.

Explanation:

To describe end behavior of a function you need to calculate the limits:

#lim_{x->-oo}f(x)#

and

#lim_{x->+oo}f(x)#

Here you get:

#lim_{x->+-oo}(8-x^3-2x^4)=lim_{x->+-oo}(x^4*(-2-1/x+8/x^3))#

If #x# goes to #+-oo# the fractions in the brackets go to zero, and #x^4# goes to #+oo#, so the whole expression goes to #-oo#. Therfore we can say that as #x# goes to #-oo# or #+oo# the function goes to #-oo#

May 28, 2018

Answer:

as #x->oo, f(x)->-oo#
as #x->-oo, f(x)->-oo#

Explanation:

#y=8-x^3-2x^4#

To determine the "end behavior" we need only consider the highest degree term because as #x->oo# the highest degree term dominates the function's behavior:

#-2x^4#

#x^4# has very similar behavior to #x^2# just grows more rapidly.

since the coefficient is negative the function's end behavior is decreasing, so we have determined:

#f(x) = -2x^4-x^3+8#

as #x->oo, f(x)->-oo#
as #x->-oo, f(x)->-oo#

graph{-2x^4-x^3+8 [-10.545, 9.455, -1.4, 8.6]}