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

Jul 10, 2015

Graph $y = x + 3 - \frac{4}{x}$ where $x \ne 0$

#### 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
$y = x + 3 - \frac{4}{x}$
(3) graph the latter as a sum of two graphs
$y = x + 3$ and $y = - \frac{4}{x}$

Graph of $y = x + 3$ is

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

Graph of $y = - \frac{4}{x}$ is

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

Summing these together, we see that around $x = 0$ the main component is $y = - \frac{4}{x}$ 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 $y = - \frac{4}{x}$ with a small corrections by adding $x + 3$.

Outside of the vicinity of point $x = 0$ main component is $y = x + 3$, while $y = - \frac{4}{x}$ 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
${x}^{2} + 3 x - 4 = 0$
Solutions are $x = 1$ and $x = - 4$.

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

As $x \to - \infty$, function $y = x + 3 - \frac{4}{x}$ 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 $+ \infty$ as $x \to 0$.

At $x = 0$ function is not defined.

As $x$ moves to a positive side, the function value grows from $- \infty$ 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]}