Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `f(x)=(5x-15) /( 2x+4)`?

Answer 1

vertical asymptote x = -2
horizontal asymptote `y=5/2`

Explanation:

The denominator of f(x) cannot be zero as this is undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: 2x + 4 = 0 → x = -2 is the asymptote

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

`((5x)/x-15/x)/((2x)/x+4/x)=(5-15/x)/(2+4/x)`

as `xto+-oo,f(x)to(5-0)/(2+0)`

`rArry=5/2" 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{(5x-15)/(2x+4) [-40, 40, -20, 20]}