Calculus

Determine the equation ofthetangent line in slopeintercept form at x=1 of y^3 +2x^2=1 at x=1. Please show work so I can follow the process. Thanks!

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  1. y^3+2x^2=1

    y^2 dy/dx+4x=0
    dy/dx=-4x/y^2

    at x=1, y=-1, so

    dy/dx=-4/1=-4
    check my thinking.

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    bobpursley
  2. Hmm. I get

    y^3 + 2x^2 = 1
    3y^2 y' + 4x = 0
    y' = -4x/(3y^2)

    y(1) = -1, so
    y'(1) = -4/3

    so, now you have a point (1,-1) and a slope: -4/3

    y+1 = -4/3 (x-1)

    I assume that massaging that to slope-intercept form presents no problems...

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