Question

Hello,solve this please.
f(x) = x^2 - ax. Find a if the tangent of f's graph is parallel with (d):y=x-2011 at M(1,f(1)).
then found and the tangent's type.

Answer 1

I'd start by thinking what is the geometric meaning of a derivative.

Answer 2

yep

Answer 3

f'(x) = 2x - a
this equals the gradient of a tangent at some point on the graph of f(x)
y = x - 2011 has a gradient of 1
so 2x-a = 1 so a = 2x-1
can u continue from here?

Answer 4

i think a must be 2011

Answer 5

why the point (1,f(1)) is given then ?