Question

How do you find the asymptotes for `y= (3(x+5))/((x+1)(x+2))`?

Answer 1

Horizontal asymptotes: `x=-1 color(white)("XX")andcolor(white)("XX")x=-2`
Vertical asymptote: `y=0`

Explanation:

The denominator of `(3(x+5))/((x+1)(x+2))` will become `0` if `x=-1` or `x=-2`
This results in horizontal asymptotes at these locations (provided the numerator does not also go to `0`, which it doesn't in this case).

Since the degree of the denominator is greater than the degree of the numerator, as `xrarroo, yrarr0`,
providing the vertical asymptote `y=0`.
This could be made explicitly clear as
`color(white)("XXX")y=(3(x+5))/((x+1)(x+2))`

`color(white)("XXX")=(3x+15)/(x^2+3x+2)`

`color(white)("XXX")=(3+15/x)/(x+3+2/x)`

As `xrarroo` `15/xrarr0` and `2/xrarr0`

leaving `lim_(xrarroo) 3/(x+3)` with the constant numerator and the constant added in the denominator becoming less significant as `xrarroo`
So `lim_(xrarroo) 3/(x+3) rarr lim_(xrarroo)1/x rarr 0`