f(x)=(x-3)(x+3)/x(x-3)= (x+3)/x
The vertical asymptote is indeed x=0
and the slanting asymptote is indeed the line y=x+3
f(x)=(x^3-27)/(x^2-3x)
I need to find the slant.
If the vertical asymptotes is x=0
then is the slant y=x+3?
2 answers
f(x)=(x-3)(x^2+3x+9)/x(x-3)=
(x^2+3x+9)/x=x+3+9/x
If x->inf then 9/x->0
(x^2+3x+9)/x=x+3+9/x
If x->inf then 9/x->0