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Gnom [1K]
2 years ago
7

HELP ASAP ...........

Mathematics
1 answer:
dedylja [7]2 years ago
8 0

Answer:

A

Step-by-step explanation:

Given

y - 2x - 8 = 0 ( add 2x + 8 to both sides )

y = 2x + 8 → (1)

y² + 8x = 0 → (2)

Substitute y = 2x + 8 into (2)

(2x + 8)² + 8x = 0 ← expand left side using FOIL and simplify

4x² + 32x + 64 + 8x = 0

4x² + 40x + 64 = 0 ( divide through by 4 )

x² + 10x + 16 = 0 ← in standard form

(x + 8)(x + 2) = 0 ← in factored form

x + 8 = 0 ⇒ x = - 8

x + 2 = 0 ⇒ x = - 2

Substitute these values into (1) for corresponding values of y

x = - 8 : y = 2(- 8) + 8 = - 16 + 8 = - 8 ⇒ P (- 8, - 8)

x = - 2 : y = 2(- 2) + 8 = - 4 + 8 = 4 ⇒ Q (- 2, 4 )

Calculate the length of PQ using the distance formula

PQ = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2    }

with (x₁, y₁ ) = P (- 8, - 8) and (x₂, y₂ ) = Q (- 2, 4 )

PQ = \sqrt{(-2-(-8))^2+(4-(-8))^2}

      = \sqrt{(-2+8)^2+(4+8)^2}

       = \sqrt{6^2+12^2}

       = \sqrt{36+144}

       = \sqrt{180}

       = \sqrt{36(5)}

       = \sqrt{36} × \sqrt{5}

        = 6\sqrt{5} → A

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Can anybody help plzz?? 65 points
Yakvenalex [24]

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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