Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 20

y=x^3-6x^2-15x+4
a) Asymptotes
Polynomial function of degree 1 or higher can't have any asymptotes.
b) Maxima/Minima
y'=3x^2-12x-15
y'=3(x^2-4x-5)
y'=3(x-5)(x+1)
Now we can find critical numbers by solving x for y'=0,
3(x-5)(x+1)=0rArrx=5 , x=-1
Now let's check the sign of y' by plugging test points in the intervals (-oo ,-1) , (-1,5) and (5,oo )
y'(-2)=3(-2-5)(-2+1)=21
y'(1)=3(1-5)(1+1)=-24
y'(6)=3(6-5)(6+1)=21
Local Maximum at x=-1
y(-1)=(-1)^3-6(-1)^2-15(-1)+4=12
Local Minimum at x=5
y(5)=5^3-6(5^2)-15(5)+4=-96
c) Inflection Points
y''=6x-12=6(x-2)
Solve for x ,y''=0
6(x-2)=0rArrx=2
y(2)=2^3-6(2^2)-15(2)+4=-42
Inflection point is (2,-42)