Question

use the limit definition of the derivative to find the slope of the tangent line to the curve y = \frac{2}{x} at the point (8, \frac{1}{4}).
the equation of this tangent line can be written in the form y = mx + b where:
m is:
b is:
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Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative of a function $y = f(x)$ is $f^\prime(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=\frac{2}{x}$, then $f(x + h)=\frac{2}{x + h}$.

Step2: Substitute into the limit - definition

$$
LATEXBLOCK0
$$

Step3: Evaluate the limit

As $h
ightarrow0$, we have $f^\prime(x)=-\frac{2}{x^{2}}$.

Step4: Find the slope $m$ at $x = 8$

Substitute $x = 8$ into $f^\prime(x)$. Then $m=f^\prime(8)=-\frac{2}{8^{2}}=-\frac{2}{64}=-\frac{1}{32}$.

Step5: Find the $y$ - intercept $b$

We know the point $(x_0,y_0)=(8,\frac{1}{4})$ and the slope $m =-\frac{1}{32}$. Using the equation $y=mx + b$, we substitute $x = 8$, $y=\frac{1}{4}$, and $m=-\frac{1}{32}$:

$$
LATEXBLOCK1
$$

Answer:

The slope of the tangent line is $-\frac{1}{32}$.
$m$ is: $-\frac{1}{32}$
$b$ is: $\frac{1}{2}$