ANSWER :
The slope is -1/8
EXPLANATION :
From the problem, we have the function :
[tex]f(x)=\frac{8}{x-3}[/tex]
The slope of the tangent line at x = -5 is the value of the first derivative at x = -5
Differentiate the function :
[tex]\begin{gathered} f(x)=\frac{8}{x-3} \\ \\ \text{ can be written as :} \\ \\ f(x)=8(x-3)^{-1} \\ \\ \text{ using the general rule for differentiation :} \\ \\ f(x)=ax^n \\ f^{\prime}(x)=n(ax^{n-1}) \\ \\ \text{ Then :} \\ \\ f^{\prime}(x)=-1[8(x-3)^{-1-1}] \\ f^{\prime}(x)=-8(x-3)^{-2} \\ \\ \text{ Write as a fraction :} \\ f^{\prime}(x)=-\frac{8}{(x-3)^2} \\ \\ \text{ Then substitute x = -5} \\ \\ f^{\prime}(-5)=-\frac{8}{(-5-3)^2}=-\frac{8}{(-8)^2} \\ \\ f^{\prime}(-5)=-\frac{8}{64}=-\frac{1}{8} \end{gathered}[/tex]
