The slope of the tangent line to a curve is another way to say the derivative at one particular value of #x#. Since it's sort of debatable where you are with this, I'll do it both the limit way and the easier way.
#f'(x) = lim_(h->0)[f(x+h) - f(x)]/h#
#= lim_(h->0)[((x+h) - (x+h)^2) - (x-x^2)]/h#
#= lim_(h->0)[(x+h) - (x^2+2hx+h^2) - (x-x^2)]/h#
#= lim_(h->0)[cancel(x) + h - cancel(x^2) - 2hx - h^2 - cancel(x) + cancel(x^2)]/h#
#= lim_(h->0)[cancel(h) - 2cancel(h)x - h^cancel(2)]/cancel(h)#
#= lim_(h->0) 1 - 2x - h#
#= 1-2x#
The easier way would have been the Power Rule:
#f(x) = x^n => f'(x) = nx^(n-1)#
#=> f'(x) = 1x^(1-1) - 2x^(2-1) = 1 - 2x#
And now to find the slope at #(1,0)#, just plug #x = 1# into #f'(x)#.
#= 1 - 2 = -1#.
graph{x-x^2 [-10, 10, -5, 5]}
