Question

How do you sketch the general shape of `f(x)=-x^4+3x^2-2-5x` using end behavior?

Answer 1

As `x to +-oo, y to -oo`. The zenith (y' = 0 ) is close to (1.5, 7.8125).
There are two points of inflexion at `x = +-sqrt2`. See the illustrative graph.

Explanation:

f(x) has alternate signs at x = 0, -1 and -2. So, the graph cuts xaxis

twice in (-2, -1).

f'=-4x^3+6x-5 and changes sign nera x = -1.5, for a turning point.

`f''=-12x^2+6= 0`, for `x= +-sqrt2 and f'''=-12`.

So, the points of inflexion are `+-(sqrt2, -5sqrt2)`.

`y=f=-x^4 (1-3/x^2+5/x^3+2/x^4) to -oo`, as `x to +-oo`.

graph{-x^4+3x^2-5x-2 [-20, 20, -10, 10]}