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Graphing Rational Functions

Graphing Rational Functions 1.7 - Solution

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To graph the given rational function, we will start by identifying its asymptotes. f(x)=x3x+1\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,x=h, where hh is the value that makes the denominator equal to zero. The horizontal asymptote is y=k,y=k, where kk is the quotient of the leading coefficients, which are both 1.1. VerticalHorizontalasymptoteasymptotex+1=0y=11x=-1y=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-x\text{-}values to the left and to the right of the horizontal asymptote.

xx x3x+1\dfrac{x-3}{x+1} f(x)=x3x+1f(x)=\dfrac{x-3}{x+1}
-6{\color{#0000FF}{\text{-} 6}} -63-6+1\dfrac{{\color{#0000FF}{\text{-} 6}}-3}{{\color{#0000FF}{\text{-} 6}}+1} 1.81.8
-4{\color{#0000FF}{\text{-} 4}} -43-4+1\dfrac{{\color{#0000FF}{\text{-} 4}}-3}{{\color{#0000FF}{\text{-} 4}}+1} 2.3\approx 2.3
-2{\color{#0000FF}{\text{-} 2}} -23-2+1\dfrac{{\color{#0000FF}{\text{-} 2}}-3}{{\color{#0000FF}{\text{-} 2}}+1} 55
0{\color{#0000FF}{0}} 030+1\dfrac{{\color{#0000FF}{0}}-3}{{\color{#0000FF}{0}}+1} -3\text{-} 3
2{\color{#0000FF}{2}} 232+1\dfrac{{\color{#0000FF}{2}}-3}{{\color{#0000FF}{2}}+1} -0.3\approx \text{-} 0.3
4{\color{#0000FF}{4}} 434+1\dfrac{{\color{#0000FF}{4}}-3}{{\color{#0000FF}{4}}+1} 0.20.2

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.