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Question 33 (3 marks) At the point (2,-5) on the curve y=f(x), the tangent has the equation 2x-y-9=0. Determine the equation of the tangent to the curve y=4-2f(3+(x)/(2)) at the point (-2,14), showing all reasoning.

Question 3333 (33 marks)\newlineAt the point (2,−5) (2,-5) on the curve y=f(x) y=f(x) , the tangent has the equation 2x−y−9=0 2 x-y-9=0 . Determine the equation of the tangent to the curve y=4−2f(3+x2) y=4-2 f\left(3+\frac{x}{2}\right) at the point (−2,14) (-2,14) , showing all reasoning.

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Q. Question 3333 (33 marks)\newlineAt the point (2,−5) (2,-5) on the curve y=f(x) y=f(x) , the tangent has the equation 2x−y−9=0 2 x-y-9=0 . Determine the equation of the tangent to the curve y=4−2f(3+x2) y=4-2 f\left(3+\frac{x}{2}\right) at the point (−2,14) (-2,14) , showing all reasoning.
  1. Determine Derivative at 2,−52, -5: Determine the derivative of f(x)f(x) using the given tangent equation at 2,−52, -5. The slope of the tangent line is the derivative of the function at that point. The equation of the tangent is 2x−y−9=02x - y - 9 = 0, rearrange to y=2x−9y = 2x - 9. The slope (derivative) at x=2x = 2 is 22.
  2. Find f′(x)f'(x) at x=2x=2: Use the derivative to find f′(x)f'(x). Since the slope of the tangent to y=f(x)y = f(x) at x=2x = 2 is 22, f′(2)=2f'(2) = 2.
  3. Derivative of y: Find the derivative of y=4−2f(3+x2)y = 4 - 2f(3 + \frac{x}{2}). Using the chain rule, dydx=−2⋅f′(3+x2)⋅(12)\frac{dy}{dx} = -2 \cdot f'(3 + \frac{x}{2}) \cdot (\frac{1}{2}). Simplify to dydx=−f′(3+x2)\frac{dy}{dx} = -f'(3 + \frac{x}{2}).
  4. Find Slope at x=−2x=-2: Substitute x=−2x = -2 into the derivative to find the slope of the tangent at that point. dydx\frac{dy}{dx} at x=−2x = -2 is −f′(3+(−2)/2)=−f′(2)=−2-f'(3 + (-2)/2) = -f'(2) = -2.
  5. Equation of Tangent at (−2,14)(-2, 14): Use the point-slope form of the equation of a line to find the equation of the tangent at (−2,14)(-2, 14). The slope is −2-2 and the point is (−2,14)(-2, 14). Equation: y−14=−2(x+2)y - 14 = -2(x + 2).
  6. Simplify Tangent Equation: Simplify the equation of the tangent. y−14=−2x−4y - 14 = -2x - 4, so y=−2x+10y = -2x + 10.

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