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

1 Answer
Dec 2, 2016

Answer:

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]}