Calculus

Evaluate slope of a tangent line:

y=1-(1/2)x^2, P(2,-1)

The answer is supposed to be -1, but I'm getting -2.

[1-2-2h-(1/2)h^2]-[-1]/h

[-1-2h-(1/2)h^2+1]/h

[h(-2-1/2h)]/h

-1/2(0)-2

=-2

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asked by Mike
  1. first using calculus to check
    y=1-(1/2)x^2, P(2,-1)
    dy/dx = -x
    at x = 2
    That is
    -2
    Like you are right, period.

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    posted by Damon
  2. y=1-(1/2)x^2
    y'=0-2/2 x
    slope= -x=-2

    The lim method:
    =1-(1/2)(x+h)^2 -1+1/2(x)^2
    = 1-1/2)(x^2+2xh+h^2)-1+1/2(x)^2 all over h
    as h > 0
    = ( -xh-h^2)/h= -x at at 2,-1
    =-2

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