Question

How do you graph `f(x) = (-3x-5)/(x+2)`?

Answer 1

`f(x)` is a rectangular hyperbola with asymtotes at `x=-2 and y=-3`
and intercepts `(0, -5/2)` and `(-5/3, 0)`

Explanation:

`f(x) = (-3x-5)/(x+2)`

`= - (3x+5)/(x+2)`

Notice `f(x)` is undefined at `x=-2`
and the domain of `f(x)` is `(-oo,-2)uu(-2,+oo)`

`f(x)` is a hyperbola.

Consider:

`lim_(x->-2^-) f(x) -> -(-6+5)/(-0) = -oo`

and

`lim_(x->-2^+) f(x) -> -(-6+5)/(+0) = +oo`

Now consider:

`lim_(x->+oo) f(x) =lim_(x->+oo) -(3+5/x)/(1+2/x) -> -(3+0)/(1+0) =-3`

and

`lim_(x->-oo) f(x) =lim_(x->-oo) -(3+5/x)/(1+2/x) -> -(3-0)/(1-0) =-3`

Hence `f(x)` is a rectangular hyperbola with asymtotes at `x=-2 and y=-3`

Other points of interest that will help in graphing `f(x)`:

`f(0) = -5/2 -> (0, -5/2)` is a point of interest

`f(x) = 0 -> x = -5/3 -> (-5/3, 0)` is a point of interest

These results help us to graph `f(x)` as below.

graph{-(3x+5)/(x+2) [-16.02, 16.02, -8.01, 8.01]}

[N.B. In practice, we may need to computs and plot a few extra points to construct the graph above.]