calculus

find an equation of the tangent line to the curve at y=2x^3+5x, (-1, 3)

  1. 0
asked by vicki
  1. dy/dx = 6x + 5
    so at (-1,3), dy/dx = -6+5 = -1

    slope = -1, point on line is (-1,3)
    3 = (-1)(-1) + b in y = mx+b
    2 = b

    equation of tangent:
    y = -x + 2

    posted by Reiny

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