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

Jul 6, 2015

Separate into a polynomial (constant) term and rational expression that tends to zero as $x \to \pm \infty$. Deduce asymptotic behaviour.

#### Explanation:

$f \left(x\right) = \frac{3 x + 3}{2 x + 4} = \frac{3 \left(x + 1\right)}{2 \left(x + 2\right)} = \frac{3}{2} \cdot \frac{x + 1}{x + 2}$

$= \frac{3}{2} \cdot \frac{x + 2 - 1}{x + 2}$

$= \frac{3}{2} \left(\frac{x + 2}{x + 2} - \frac{1}{x + 2}\right)$

$= \frac{3}{2} \left(1 - \frac{1}{x + 2}\right)$

$= \frac{3}{2} - \frac{3}{2 \left(x + 2\right)}$

with exclusion $x \ne - 2$

So as $x \to - \infty$ the term $\frac{3}{2 \left(x + 2\right)} \to {0}_{-}$ so $f \left(x\right) \to {\left(\frac{3}{2}\right)}_{+}$

As $x \to {\left(- 2\right)}_{-}$ the term $\frac{3}{2 \left(x + 2\right)} \to - \infty$ so $f \left(x\right) \to \infty$

As $x \to {\left(- 2\right)}_{+}$ the term $\frac{3}{2 \left(x + 2\right)} \to \infty$ so $f \left(x\right) \to - \infty$

As $x \to \infty$ the term $\frac{3}{2 \left(x + 2\right)} \to {0}_{+}$ so $f \left(x\right) \to {\left(\frac{3}{2}\right)}_{-}$

The intersection with the $x$ axis occurs at $\left(- 1 , 0\right)$ since the numerator of the original expression is $0$ for $x = - 1$

The intersection with the $y$ axis can be found by substituting $x = 0$ into the original equation to derive $\left(0 , \frac{3}{4}\right)$

graph{ (3x+3)/(2x+4) [-12.29, 7.71, -3.44, 6.56]}