By definition of the derivative, and with [tex]f(x)=-x^2+4x-9[/tex], we have
[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h[/tex]
In this case,
[tex]f(x+h)=-(x+h)^2+4(x+h)-9=-x^2-2xh-h^2+4x+4h-9=f(x)+(4-2x)h-h^2[/tex]
Then
[tex]f(x+h)-f(x)=(4-2x)h-h^2[/tex]
[tex]\implies\dfrac{f(x+h)-f(x)}h=4-2x-h[/tex]
so that as [tex]h\to0[/tex], we're left with
[tex]f'(x)=4-2x[/tex]
and so
[tex]f'(1)=2[/tex]
[tex]f'(2)=0[/tex]
[tex]f'(3)=-2[/tex]
