# How do you graph  f(x)= x/(x+3)?

Jul 28, 2015

graph{1-3/(x+3) [-10, 10, -15, 15]}

#### Explanation:

First of all, we see that the function is not defined at $x = - 3$. In fact, there is a vertical asymptote at this point since the denominator tends to $0$ as $x \to - 3$, while the numerator is around a constant $- 3$.

More than that, as $x \to - 3$ from the right (that is, while it's greater than $- 3$), the sign of the function is negative since the numerator is around $- 3$ and denominator is positive, so the function tends to negative infinity $- \infty$.
If $x \to - 3$ from the left (that is, while it's smaller than $- 3$), the sign of the function is positive since the numerator is around $- 3$ and denominator is negative, so the function tends to positive infinity $+ \infty$.

Continuing analysis, we can see that, as $x \to + \infty$ or as $x \to - \infty$, the value of the function tends to $1$ since both numerator and denominator will increase to $+ \infty$ or decrease to $- \infty$ with the same speed.

The easiest way is to transform our function as follows:
$y = \frac{x}{x + 3} = \frac{x + 3 - 3}{x + 3} = \frac{x + 3}{x + 3} - \frac{3}{x + 3} = 1 - \frac{3}{x + 3}$

So, we have to graph $y = 1 - \frac{3}{x + 3}$.
According to the rules of graph transformation (seeUnizor - Algebra - Graph) we can construct this graph in the following steps:

Step 1. Graph $y = \frac{1}{x}$

graph{1/x [-10, 10, -15, 15]}

Step 2. Shift it to the left by $3$ to graph $y = \frac{1}{x + 3}$

graph{1/(x+3) [-10, 10, -15, 15]}

Step 3. Stretch it vertically by a factor of $3$ getting the graph of $y = \frac{3}{x + 3}$

graph{3/(x+3) [-10, 10, -15, 15]}

Step 4. Invert the graph (positive - to negative, negative - to positive), thus getting the graph of $y = - \frac{3}{x + 3}$

graph{-3/(x+3) [-10, 10, -15, 15]}

Step 5. Finally, shift the graph up by $1$ to get $y = 1 - \frac{3}{x + 3}$:

graph{1-3/(x+3) [-10, 10, -15, 15]}