## Calculus with Applications (10th Edition)

$y = - 2x + 9$
$\begin{gathered} f\,\left( x \right) = \frac{x}{{x - 2}} \hfill \\ Use\,\,the\,\,quotietn\,\,rule\,\,to\,\,find\,\,{f^,}\,\left( x \right) \hfill \\ {f^,}\,\left( x \right) = \frac{{\,\left( {x - 2} \right)\,{{\left( x \right)}^,} - \,\left( x \right)\,{{\left( {x - 2} \right)}^,}}}{{\,{{\left( {x - 2} \right)}^2}}} \hfill \\ Then \hfill \\ {f^,}\,\left( x \right) = \frac{{\,\left( {x - 2} \right)\,\left( 1 \right) - \,\left( x \right)\,\left( 1 \right)}}{{\,{{\left( {x - 2} \right)}^2}}} \hfill \\ Simplifying \hfill \\ {f^,}\,\left( x \right) = \frac{{x - 2 - x}}{{\,{{\left( {x - 2} \right)}^2}}} = - \frac{2}{{\,{{\left( {x - 2} \right)}^2}}} \hfill \\ Evaluate\,\,{f^,}\,\left( x \right)\,\,at\,\,the\,\,point\,\,\,\left( {3,3} \right) \hfill \\ {f^,}\,\left( 3 \right) = - \frac{2}{{\,{{\left( {3 - 2} \right)}^2}}} = - 2 \hfill \\ Then \hfill \\ m = - 2 \hfill \\ Use\,\,y - {y_1} = m\,\left( {x - {x_1}} \right) \hfill \\ y - 3 = - 2\,\left( {x - 3} \right) \hfill \\ Simplifying \hfill \\ y - 3 = - 2x + 6 \hfill \\ y = - 2x + 9 \hfill \\ \end{gathered}$