fckm

the slope of the tangent line at the point is equal to the derivative at the point.
for:
y=1=2x-x^3,

dy/dx= 2-3x^2

evaluate at x=1
dy/dx=-1

the "slope-intercept" form of a line is:

y=mx+b

we found m=-1

and we need this line to go through the point (1,2)

plug in y=1, x=1:

2=-1+b

gets you b=3.

therefore, the tangent line has the form:

y=-1x+3

The formula that you have, (f(x) - f(a)) / (x - a), is sorta like an average slope. To get the instantaneous slope at a point, you must take the limit as x->a. Your formula then becomes the definition of a derivative (at a point x=a).

Rereading your post, you probably know this already. To take the limit, you either factor (i can't remember how to factor cubics right now, use the SOAP rule), or just use L'Hopital's rule.

EDIT: Matlab says 2x-x^3-1 = -(x-1)(x^2+x-1).
So you were on the right track. after you factor this and cancel with the bottom, plugging in x=1, you get a slope of -1.

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