We're given that
[tex]f(x)=x^2-2x+6[/tex]
which means
[tex]f(x+h)=(x+h)^2-2(x+h)+6=x^2+2xh+h^2-2x-2h+6[/tex]
When we subtract these, several terms cancel:
[tex](x^2+2xh+h^2-2x-2h+6)-(x^2-2x+6)=2xh+h^2-2h[/tex]
So the difference quotient is
[tex]\dfrac{f(x+h)-f(x)}h=\dfrac{2xh+h^2-2h}h[/tex]
and since h ≠ 0, we can cancel it out to end up with
[tex]\dfrac{f(x+h)-f(x)}h=2x+h-2[/tex]
Now plug in x = 8 (this could have been done at any prior point):
[tex]\dfrac{f(8+h)-f(8)}h=14+h[/tex]
