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March 26, 2017

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Determine the slope of the tangent to the curve y=2(x^2+x-1)^3 at (-1, -2)

  • Calculus - ,

    dy/dx= 6(x^2+x-1)^2 (2x+1)
    at (-1,-2)

    dy/dx= 6(1-1-1)^2 (-2+1)
    = -6

    so now you have the slope of -6 and a point(-1,-2)

    continue ....

  • Calculus - ,

    So

    y2-y1=m(x2-x1)

    y-(-2)=-6(x-(-1)
    y=6x-1

    therefore the slope of tangent is 6?

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