Question

How do you graph `(x^2+3x-4)/x`?

Answer 1

Graph `y=x+3-4/x` where `x != 0`

Explanation:

A relatively easy approach is to
(1) notice that the function is not defined at `x=0` since it contains `x` in the denominator
(2) at every other point in its domain the numerator can be divided by denominator getting
`y=x+3-4/x`
(3) graph the latter as a sum of two graphs
`y=x+3` and `y=-4/x`

Graph of `y=x+3` is

graph{x+3 [-10, 10, -5, 5]}

Graph of `y=-4/x` is

graph{-4/x [-10, 10, -5, 5]}

Summing these together, we see that around `x=0` the main component is `y=-4/x` since its value is infinitely large by absolute value (positive at `x < 0` and negative at `x > 0`). So, the graph looks pretty much like `y=-4/x` with a small corrections by adding `x+3`.

Outside of the vicinity of point `x=0` main component is `y=x+3`, while `y=-4/x` brings just a small corrections.

All we have to find to position the graph more precisely is to find where it crosses the X-axis, that is where the function equals to zero.
This can be accomplished by solving a quadratic equation
`x^2+3x-4=0`
Solutions are `x=1` and `x=-4`.

That leads us to describe the behavior of the original function as follows.

As `x->-oo`, function `y=x+3-4/x` looks close to a straight line `y=x+3`.

As `x` approach to point `-4`, our function deviates from the straight line behavior and starts growing faster and faster.

After crossing the X-axis at point `x=-4` the function asymptotically increases to `+oo` as `x->0`.

At `x=0` function is not defined.

As `x` moves to a positive side, the function value grows from `-oo` to `0` at `x=1`.

Then the growth gradually becomes more linear and eventually follows approximately the straight line `y=x+3`

The final graph looks like this

graph{x+3-4/x [-20, 20, -25, 25]}