Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 25

y=x^2/(x+8)
a) Asymptotes
Vertical asymptotes are the zeros of the denominator
x+8=0rArrx=-8
Vertical asymptote is x=-8
Degree of numerator=2
Degree of denominator=1
Degree of numerator=1+Degree of denominator, so the asymptote is a slant asymptote of the form y=mx+b
For a rational function the slant asymptote is the quotient of the polynomial division.
x^2/(x+8)=x-8+64/(x+8)
the slant asymptote is y=x-8
b) Maxima/Minima
y'=((x+8)2x-x^2)/(x+8)^2
y'=(2x^2+16x-x^2)/(x+8)^2
y'=(x(x+16))/(x+8)^2
Let's find critical numbers by solving x for y'=0,
x(x+16)=0rArrx=0 , x=-16
y(-16)=(-16)^2/(-16+8)=-32
y(0)=0^2/(0+8)=0
Local maximum=-32 at x=-16
Local minimum=0 at x=0
c) Inflection points
y''=((x+8)^2(2x+16)-(x^2+16x)(2)(x+8))/(x+8)^4

y''=(2x^2+16x+16x+128-2x^2-32x)/(x+8)^3
y''=128/(x+8)^3
128/(x+8)^3=0
there is no solution for x , so there are no inflection points.