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Ksivusya [100]
3 years ago
15

Holly has a budget of $40 per month to spend on movies. She made a table showing the total amount of movie money she will have l

eft after seeing the given number of movies. Which equation, in Slope Intercept form, represents the relationship between the number of movies, x, and the amount of movie money Holly will have left, y?
Mathematics
1 answer:
sammy [17]3 years ago
3 0
The slope intercept equation is y=mx + b

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Can anybody help plzz?? 65 points
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Answer:

\frac{dy}{dx} =\frac{-8}{x^2} +2

\frac{d^2y}{dx^2} =\frac{16}{x^3}

Stationary Points: See below.

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Equality Properties

<u>Calculus</u>

Derivative Notation dy/dx

Derivative of a Constant equals 0.

Stationary Points are where the derivative is equal to 0.

  • 1st Derivative Test - Tells us if the function f(x) has relative max or mins. Critical Numbers occur when f'(x) = 0 or f'(x) = undef
  • 2nd Derivative Test - Tells us the function f(x)'s concavity behavior. Possible Points of Inflection/Points of Inflection occur when f"(x) = 0 or f"(x) = undef

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>

f(x)=\frac{8}{x} +2x

<u>Step 2: Find 1st Derivative (dy/dx)</u>

  1. Quotient Rule [Basic Power]:                    f'(x)=\frac{0(x)-1(8)}{x^2} +2x
  2. Simplify:                                                      f'(x)=\frac{-8}{x^2} +2x
  3. Basic Power Rule:                                     f'(x)=\frac{-8}{x^2} +1 \cdot 2x^{1-1}
  4. Simplify:                                                     f'(x)=\frac{-8}{x^2} +2

<u>Step 3: 1st Derivative Test</u>

  1. Set 1st Derivative equal to 0:                    0=\frac{-8}{x^2} +2
  2. Subtract 2 on both sides:                         -2=\frac{-8}{x^2}
  3. Multiply x² on both sides:                         -2x^2=-8
  4. Divide -2 on both sides:                           x^2=4
  5. Square root both sides:                            x= \pm 2

Our Critical Points (stationary points for rel max/min) are -2 and 2.

<u>Step 4: Find 2nd Derivative (d²y/dx²)</u>

  1. Define:                                                      f'(x)=\frac{-8}{x^2} +2
  2. Quotient Rule [Basic Power]:                  f''(x)=\frac{0(x^2)-2x(-8)}{(x^2)^2} +2
  3. Simplify:                                                    f''(x)=\frac{16}{x^3} +2
  4. Basic Power Rule:                                    f''(x)=\frac{16}{x^3}

<u>Step 5: 2nd Derivative Test</u>

  1. Set 2nd Derivative equal to 0:                    0=\frac{16}{x^3}
  2. Solve for <em>x</em>:                                                    x = 0

Our Possible Point of Inflection (stationary points for concavity) is 0.

<u>Step 6: Find coordinates</u>

<em>Plug in the C.N and P.P.I into f(x) to find coordinate points.</em>

x = -2

  1. Substitute:                    f(-2)=\frac{8}{-2} +2(-2)
  2. Divide/Multiply:            f(-2)=-4-4
  3. Subtract:                       f(-2)=-8

x = 2

  1. Substitute:                    f(2)=\frac{8}{2} +2(2)
  2. Divide/Multiply:            f(2)=4 +4
  3. Add:                              f(2)=8

x = 0

  1. Substitute:                    f(0)=\frac{8}{0} +2(0)
  2. Evaluate:                      f(0)=\text{unde} \text{fined}

<u>Step 7: Identify Behavior</u>

<em>See Attachment.</em>

Point (-2, -8) is a relative max because f'(x) changes signs from + to -.

Point (2, 8) is a relative min because f'(x) changes signs from - to +.

When x = 0, there is a concavity change because f"(x) changes signs from - to +.

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