Question

How do you find vertical, horizontal and oblique asymptotes for `f(x)= (x^2-5x+6)/ (x^2-8x+15)`?

Answer 1

vertical asymptote x = 5
horizontal asymptote y = 1

Explanation:

The first step here is to factorise and simplify f(x).

`rArrf(x)=(cancel((x-3))(x-2))/(cancel((x-3))(x-5))=(x-2)/(x-5)`

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) to 0`

divide terms on numerator/denominator by x.

`rArr(x/x-2/x)/(x/x-5/x)=(1-2/x)/(1-5/x)`

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

`rArry=1" 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{(x-2)/(x-5) [-20, 20, -10, 10]}