Question

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

Answer 1

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`

Answer 2

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]}