Find an equation for the tangent line to the curve given parametrically by y(t) = 4t² + t - 1 2t x(t) = e²t, at the point P(1, −1). 1 3 2 1. y = X 2 2. Y 1 = --x- 2 3. y = x - 4. y = x 5. y = 6. y = 12 32 1 2 1 2 -X- X 1|2 1 2 312 Find the x-intercept of the tangent line at P(4, 1) to the graph of the curve defined parametrically by x(t) = 8 cos 2t, y(t) = 2 sint. x-intercept 1. 2. x-intercept 3. x-intercept 4. x-intercept = = 12 - - -6 -3 -12 1

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A: We will find the equation of the tangent line to the curve at the given point as following.

Q: Find the equation of the tangent line at (1,2) for x'+2y = 9 y- A= B(x - C)

A: point (1,2)  ; x3+2y2=9 In the given point x is 1, y is 2

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Q: 8. Find the equation of the horizontal tangent line to the curve y = -x + yx.

A: For horizontal  tangent line put the derivative  of function equals to zero

Q: Eliminate the parameter t to find a Cartesian equation of a curve. x = 1 + 3t,  y = 2-t^2

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