use the fact that the derivative of the function f(x)=5/x is f(x)=-5/x^2 to find the equation of the tangent…

Explanation:

Step1: Find the slope of the tangent line

The derivative $f'(x)$ gives the slope of the tangent - line. Substitute $x = - 6$ into $f'(x)=\frac{-5}{x^{2}}$. Then $m=f'(-6)=\frac{-5}{(-6)^{2}}=-\frac{5}{36}$.

Step2: Find the y - coordinate of the point on the function

Substitute $x=-6$ into $f(x)=\frac{5}{x}$. So $y = f(-6)=\frac{5}{-6}=-\frac{5}{6}$.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-6,-\frac{5}{6})$ and $m = -\frac{5}{36}$. [ \begin{align*} y+\frac{5}{6}&=-\frac{5}{36}(x + 6)\ y+\frac{5}{6}&=-\frac{5}{36}x-\frac{5}{6}\ y&=-\frac{5}{36}x-\frac{5}{3}-\frac{5}{6}+\frac{5}{6}\ y&=-\frac{5}{36}x-\frac{5}{3} \end{align*} ]

Answer:

$y =-\frac{5}{36}x-\frac{5}{3}$