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# Graphing Rational Functions

## Graphing Rational Functions 1.7 - Solution

To graph the given rational function, we will start by identifying its asymptotes. $\begin{gathered} f(x)=\dfrac{x-3}{x+1} \end{gathered}$ Recall that division by zero is not defined. This means the denominator cannot be zero. Therefore, the vertical asymptote is $x=h,$ where $h$ is the value that makes the denominator equal to zero. The horizontal asymptote is $y=k,$ where $k$ is the quotient of the leading coefficients, which are both $1.$ $\begin{array}{c|c} \textbf{Vertical} \quad & \quad \textbf{Horizontal} \\ \textbf{asymptote} \quad & \quad \textbf{asymptote} \\[0.8em] x+1=0 \quad & \quad y=\dfrac{1}{1}\\ \Updownarrow \quad & \quad \Updownarrow \\ x=\text{-} 1 \quad & \quad y=1 \end{array}$ Let's now find some points on the graph. Make sure to use $x\text{-}$values to the left and to the right of the horizontal asymptote.

$x$ $\dfrac{x-3}{x+1}$ $f(x)=\dfrac{x-3}{x+1}$
${\color{#0000FF}{\text{-} 6}}$ $\dfrac{{\color{#0000FF}{\text{-} 6}}-3}{{\color{#0000FF}{\text{-} 6}}+1}$ $1.8$
${\color{#0000FF}{\text{-} 4}}$ $\dfrac{{\color{#0000FF}{\text{-} 4}}-3}{{\color{#0000FF}{\text{-} 4}}+1}$ $\approx 2.3$
${\color{#0000FF}{\text{-} 2}}$ $\dfrac{{\color{#0000FF}{\text{-} 2}}-3}{{\color{#0000FF}{\text{-} 2}}+1}$ $5$
${\color{#0000FF}{0}}$ $\dfrac{{\color{#0000FF}{0}}-3}{{\color{#0000FF}{0}}+1}$ $\text{-} 3$
${\color{#0000FF}{2}}$ $\dfrac{{\color{#0000FF}{2}}-3}{{\color{#0000FF}{2}}+1}$ $\approx \text{-} 0.3$
${\color{#0000FF}{4}}$ $\dfrac{{\color{#0000FF}{4}}-3}{{\color{#0000FF}{4}}+1}$ $0.2$

Finally, we will plot and connect the obtained points. Do not forget to graph the asymptotes, and recall that a rational function has two branches.