graph the rational function.\n f(x)=\frac{-x^{2}+4x + 4}{x - 3} \nstart by drawing the asymptotes. then plot…

Explanation:

Step1: Find the vertical asymptote

Set the denominator equal to 0. So, $x - 3=0$, which gives $x = 3$.

Step2: Find the slant asymptote

Perform polynomial long - division of $-x^{2}+4x + 4$ by $x - 3$. $-x^{2}+4x + 4=(-x + 1)(x - 3)+7$. So the slant asymptote is $y=-x + 1$.

Step3: Find points on the left - hand side of the vertical asymptote

Let $x = 2$, then $f(2)=\frac{-2^{2}+4\times2 + 4}{2 - 3}=\frac{-4 + 8+4}{-1}=-8$. Let $x = 1$, then $f(1)=\frac{-1^{2}+4\times1 + 4}{1 - 3}=\frac{-1 + 4+4}{-2}=-\frac{7}{2}$.

Step4: Find points on the right - hand side of the vertical asymptote

Let $x = 4$, then $f(4)=\frac{-4^{2}+4\times4 + 4}{4 - 3}=4$. Let $x = 5$, then $f(5)=\frac{-5^{2}+4\times5 + 4}{5 - 3}=\frac{-25 + 20+4}{2}=-\frac{1}{2}$.

To graph: Draw the vertical asymptote $x = 3$ and the slant asymptote $y=-x + 1$ as dashed lines. Plot the points $(2,-8),(1,-\frac{7}{2})$ on the left - hand side of $x = 3$ and $(4,4),(5,-\frac{1}{2})$ on the right - hand side of $x = 3$ and then sketch the graph of the rational function approaching the asymptotes.

Answer:

Graph with vertical asymptote $x = 3$, slant asymptote $y=-x + 1$ and points $(2,-8),(1,-\frac{7}{2}),(4,4),(5,-\frac{1}{2})$ plotted.