Let [tex]f(x)=-2x+1[/tex]. By definition, the derivative is
[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h[/tex]
So
[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{(-2(x+h)+1)-(-2x+1)}h=\lim_{h\to0}\frac{-2h}h=\lim_{h\to0}-2=-2[/tex]
and the slope of any line tangent to [tex]f(x)[/tex] is -2.
