use the fact that the derivative of the function f(x) = 1/x is f(x)= -1/x^2 to find the equation of the…

Explanation:

Step1: Find the slope of the tangent line

The derivative $f'(x)$ gives the slope of the tangent line. Substitute $x = - 6$ into $f'(x)=\frac{-1}{x^{2}}$. $f'(-6)=\frac{-1}{(-6)^{2}}=-\frac{1}{36}$

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

Substitute $x=-6$ into $f(x)=\frac{1}{x}$. $f(-6)=\frac{1}{-6}=-\frac{1}{6}$

Step3: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)$ is the point on the curve and $m$ is the slope. Here $x_1=-6,y_1 =-\frac{1}{6},m =-\frac{1}{36}$. $y-(-\frac{1}{6})=-\frac{1}{36}(x - (-6))$ $y+\frac{1}{6}=-\frac{1}{36}(x + 6)$ $y+\frac{1}{6}=-\frac{1}{36}x-\frac{1}{6}$ $y=-\frac{1}{36}x-\frac{1}{3}$

Answer:

$y =-\frac{1}{36}x-\frac{1}{3}$