Question

How do you find vertical, horizontal and oblique asymptotes for `f(x)= (5x-15)/(2x+4)`?

Answer 1

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

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

Horizontal asymptotes occur as `lim_(x to +-oo) f(x) to 0`

divide terms on numerator/denominator by x

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

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

`rArr y = 5/2 " is the asymptote "`

Oblique asymptotes occur when the degree of the numerator > degree of denominator. This is not the case here hence there are no oblique asymptotes.
graph{(5x-15)/(2x+4) [-14.24, 14.24, -7.11, 7.13]}