Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 19

You need to evaluate the horizontal asymptotes to the graph of function, such that:
lim_(x->+oo) 2-2x-x^3 = oo
lim_(x->-oo) 2-2x-x^3 = oo
Notice that the graph has no vertical or horizontal asymptotes. You need to check if there are slant asymtpotes, y = ax+b, such that:
a = lim_(x->oo) (f(x))/x = lim_(x->oo) (2-2x-x^3)/x = oo
Since a ->oo, yields that there are no slant asymptotes.
You need to determine the extremes of the functions, hence, you need to evaluate the zeroes of derivative of function, such that:
f'(x) = -2 - 3x^2 =>f'(x) = 0=> -2 - 3x^2 = 0=> 3x^2 = -2 => x^2 = -2/3 => x_(1,2) !in R
Hence, there are no absolute extrema of the function.
You need to determine the inflection points, hence, you need to find the zeroes of the second derivative:
f''(x) = -6x => f''(x) = 0 => -6x = 0 => x = 0 => f(0) = 2.
Hence, the function has an inflection point (0,2).
The graph of the function f(x) = 2-2x-x^3 , whose inflection point,where the graph changes its curvature, is at (0,2), is represented below.