1. use the definition of derivative to find an equation of the line tangent to the curve y = 1/(x + 9) at…

Explanation:

Step1: Recall derivative definition

The definition of the derivative of a function $y = f(x)$ is $f^\prime(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$. Here $f(x)=\frac{1}{x + 9}$, so $f(x+h)=\frac{1}{(x + h)+9}$. Then $\frac{f(x + h)-f(x)}{h}=\frac{\frac{1}{x + h+9}-\frac{1}{x + 9}}{h}$.

Step2: Simplify the difference - quotient

First, find a common denominator for the numerator: $\frac{1}{x + h+9}-\frac{1}{x + 9}=\frac{(x + 9)-(x + h + 9)}{(x + h+9)(x + 9)}=\frac{x + 9-x - h - 9}{(x + h+9)(x + 9)}=\frac{-h}{(x + h+9)(x + 9)}$. So $\frac{\frac{1}{x + h+9}-\frac{1}{x + 9}}{h}=\frac{\frac{-h}{(x + h+9)(x + 9)}}{h}=\frac{-1}{(x + h+9)(x + 9)}$.

Step3: Find the derivative

Take the limit as $h\rightarrow0$: $f^\prime(x)=\lim_{h\rightarrow0}\frac{-1}{(x + h+9)(x + 9)}=-\frac{1}{(x + 9)^2}$.

Step4: Find the slope of the tangent line at $x=-8$

Substitute $x = - 8$ into the derivative. $f^\prime(-8)=-\frac{1}{(-8 + 9)^2}=-1$.

Step5: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-8,1)$ and $m=-1$. Substituting these values, we get $y - 1=-1(x+8)$.

Step6: Simplify the equation of the tangent line

Expand the right - hand side: $y - 1=-x - 8$. Then $y=-x - 7$.

Answer:

$y=-x - 7$