Question

How do you find the horizontal asymptote for `fR(x)=(3x+5) /(x-6)`?

Answer 1

The horizontal asymptote is `y=3`

Explanation:

Calculate the limit as the function `f(x)` tend to `+oo` and `-oo`

The function is

`f(x)=(3x+5)/(x-6)`

The domain of `f(x)` is `RR-{6}`

`f(x)=(3x+5)/(x-6)=(cancelx(3+5/x))/(cancelx(1-6/x))=(3+5/x)/(1-6/x)`

The limits are

`lim_(x->+oo)(5/x)=0`

`lim_(x->+oo)(6/x)=0`

`lim_(x->-oo)(5/x)=0`

`lim_(x->-oo)(6/x)=0`

And finally

`lim_(x->+oo)f(x)=lim_(x->+oo)(3+5/x)/(1-6/x)=(3-0)/(1-0)=3`

The horizontal asymptote is `y=3`

graph{(y-(3x+5)/(x-6))(y-3)=0 [-34.77, 47.43, -16.94, 24.16]}