In Problems $7-44$, follow Steps $1$through $7$on page $228$to analyze the graph of each function.

$R\left(x\right)=\frac{3x+3}{2x+4}$

$R\left(x\right)=\frac{3x+3}{2x+4}$

which may also written as,

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

The domain of $R$is, $\left\{x|x\ne -2\right\}$.

As $0$is in the domain, substitute $x=0$in the given equation.

$$\begin{gathered} R\left(0\right)=\frac{0+3}{0+4} \\ =\frac{3}{4} \end{gathered}$$

So, the obtained point is,$\left(0,\frac{3}{4}\right)$.

The real zeros of the numerator satisfies the equation $x+1=0$.

So, the $x$- intercept is,$-1$.

The real zeros of the numerator are the real solutions of the equation

$$\begin{gathered} 2\left(x+2\right)=0 \\ x+2=0 \\ x=-2 \end{gathered}$$

So, the line $x=-2$is the vertical asymptote of the graph.

Degree of the numerator is equal to the degree of the denominator. So, $R$is improper.

Thus, the graph has a horizontal asymptote. The leading coefficient of the numerator is $3$and that of the denominator is $2$. The graph of$R$has a horizontal asymptote ofrole="math" localid="1646230515396" $y=\frac{3}{2}$

Substitute $R\left(x\right)=\frac{3}{2}$in the given function.

$$\begin{gathered} \frac{3}{2}=\frac{3(x+1)}{2(x+2)} \\ 1=\frac{\left(x+1\right)}{\left(x+2\right)} \\ (x+2)=(x+1) \\ 2\ne 1 \end{gathered}$$

So, there is no solution for $x$and the graph does not intersect the horizontal asymptote $y=\frac{3}{2}$at any of the points.