# 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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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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