Find All Points Of Horizontal And Vertical...

Anonymous

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Find all points of horizontal and vertical tangency to thecurve.graph the problem x=t+2, y=t^3-2t

 

Answer No.1

Anonymous

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Dy/dx = 0 for horizontal tangent. dy/dx = (dy/dt )/(dx/dt) dy/dt = 3t^2 - 2 = 3*(x-2)^2 - 2 dx/dt = 1 dy/dt / dx/dt = dy/dx = (3*(x-2)^2)-2)/1 Thisderivative never equals inf, so no vertical tangents (slope =inifinity) So we set dy/dx = 0 for horizontal tangents 3*(x-2)^2 - 2 = 0 3*x^2 -12x + 12 - 2 = 0 3x^2 - 12x + 10 = 0 x = {12 +/- sqrt(144 - 4*3(10))}/(2*3) x = {12 +/- sqrt(24)}/6 x = 2 +/- sqrt(6)/3 At x = 2 + sqrt(6)/3, t = sqrt(6)/3 So y = 6*sqrt(6)/27 - 2*sqrt(6)/3 = -4*sqrt(6)/9 (x,y) = (2 + sqrt(6)/3, -4*sqrt(6)/9) At x = 2 - sqrt(6)/3; t = -sqrt(6)/3 y = -6*sqrt(6)/27 + 2*sqrt(6)/3 = 4*sqrt(6)/9 (x,y) = (2 - sqrt(6)/3, 4*sqrt(6)/9) graph is y = (x-2)^3 - 2*(x-2)

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