Question

the slope of the tangent line to the curve ( y = \frac{2}{x} ) at the point (2,1) is:
box
the equation of this tangent line can be written in the form ( y = mx + b ) where:
m is: box
b is: box

Explanation:

Step1: Find the derivative of \( y = \frac{2}{x} \)

Rewrite \( y = \frac{2}{x} \) as \( y = 2x^{-1} \). Using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \), the derivative \( y'=-2x^{-2}=-\frac{2}{x^{2}} \).

Step2: Find the slope at \( x = 2 \)

Substitute \( x = 2 \) into the derivative. \( y'(2)=-\frac{2}{2^{2}}=-\frac{2}{4}=-\frac{1}{2} \). So the slope \( m = -\frac{1}{2} \).

Step3: Find the equation of the tangent line

Use the point - slope form \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(2,1) \) and \( m = -\frac{1}{2} \).
\( y - 1=-\frac{1}{2}(x - 2) \)
Expand the right - hand side: \( y - 1=-\frac{1}{2}x + 1 \)
Add 1 to both sides: \( y=-\frac{1}{2}x+2 \). So \( b = 2 \).

Answer:

The slope of the tangent line: \(-\frac{1}{2}\)
\(m\) is: \(-\frac{1}{2}\)
\(b\) is: \(2\)