Question

At the indicated point find (a) the slope of the curve and (b) an equation of the tangent line.

Answer 1

Answer:

Slope of the Curve: $f'(x)=\frac{-6}{x^2}$

Equation of Tangent Line: y + 3 = -3/2(x + 2)

General Formulas and Concepts:

Pre-Algebra

• Order of Operations: BPEMDAS

Algebra I

Point-Slope Form: y - y₁ = m(x - x₁)  

• x₁ - x coordinate
• y₁ - y coordinate
• m - slope

Calculus

The definition of a derivative is the slope of the tangent line.

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

$f(x)=\frac{6}{x}$

Step 2: Take Derivative

1. Quotient Rule:                    $f'(x)=\frac{0(x)-6(1)}{x^2}$
2. Multiply:                              $f'(x)=\frac{0-6}{x^2}$
3. Subtract:                             $f'(x)=\frac{-6}{x^2}$

Step 3: Find Instantaneous Derivative

1. Substitute in x:                     $f'(x)=\frac{-6}{(-2)^2}$
2. Exponents:                           $f'(x)=\frac{-6}{4}$
3. Simplify:                               $f'(x)=\frac{-3}{2}$

This value shows the slope of the tangent line at the exact value of x = 2.

1. Substitute:                    y + 3 = -3/2(x + 2)