Question

How do you find vertical, horizontal and oblique asymptotes for `(2x+3)/(x-5)`?

Answer 1

vertical asymptote x = 5
horizontal asymptote y = 2

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve : x - 5 = 0 → x = 5 is the asymptote

Horizontal asymptotes occur as

`lim_(xto+-oo),f(x)toc" (a constant)"`

divide numerator/denominator by x

`((2x)/x+3/x)/(x/x-5/x)=(2+3/x)/(1-5/x)`

as `xto+-oo,f(x)to(2+0)/(1-0)`

`rArry=2" is the asymptote"`

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1). Hence there are no oblique asymptotes.
graph{(2x+3)/(x-5) [-20, 20, -10, 10]}