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kirza4 [7]
3 years ago
5

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

Mathematics
1 answer:
Marianna [84]3 years ago
5 0

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:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS

<u>Algebra I</u>

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

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

<u>Calculus</u>

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:

<u>Step 1: Define</u>

<u />f(x)=\frac{6}{x}<u />

<u />

<u>Step 2: Take Derivative</u>

  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}

<u>Step 3: Find Instantaneous Derivative</u>

  1. Substitute in <em>x</em>:                     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)
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