please answer with steps Normal to a curve Find an equation for the line perpend

Question

please answer with steps

Normal to a curve Find an equation for the line perpendicular to the tangent to the curve y = x3 - 4x + 1 at the point (2,1). Smallest slope What is the smallest slope on the curve? At what point on the curve does the curve have this slope? Tangents having specified slope Find equations for the tangents to the curve at the points where the slope of the curve is 8.

Explanation / Answer

a)y = x^3-4x+1 Slope of tangent = m = dy/dx = 3x^2-4 = (3*4)-4 = 12-4 = 8 Slope of line perpendicular to tangent = -1/m = -1/8 So equation of line perpendicular to tangent is given by, y - 1 = (-1/8)*(x-2) =>8y-8 = -x+2 =>x+8y = 10 b)Slope = m = 3x^2-4 To find smallest slope, differentiate m. dm/dx = 6x = 0 => x = 0 So smalles m = -4 Point on the curve is (x,y) = (0,1) c)Slope = 8 =>3x^2-4 = 8 =>3x^2 = 12 =>x^2 = 4 =>x = 2, -2 =>y = 1, 1 Equation of tangent at (2,1) is: y-1 = 8(x-2) =>y - 1 = 8x - 16 =>8x - y = 15 Equation of tangent at (-2,1) is: y-1 = 8(x+2) =>y - 1 = 8x + 16 =>8x - y = -17

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