Given f(x)=4+3x-x^2, then

f(x+h)=4+3(x+h)-(x+h)^2.

Consider the difference f(x+h)-f(x):

-(x+h)^2+x^2 + 3x+3h-3x + 4-4 =

-x^2-2xh-h^2+x^2 + 3h =

(3-2x)*h - h^2.

This divided by h is equal to

3-2x-h.

For x=3 as asked it is -3-h.

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Given f(x)=4+3x-x^2, then

f(x+h)=4+3(x+h)-(x+h)^2.

Consider the difference f(x+h)-f(x):

-(x+h)^2+x^2 + 3x+3h-3x + 4-4 =

-x^2-2xh-h^2+x^2 + 3h =

(3-2x)*h - h^2.

This divided by h is equal to

3-2x-h.

For x=3 as asked it is -3-h.

Given the function `f(x)=4+3x-x^2`

and the difference quotient `(f(3+h)-f(3))/(h)`

`(4+3(3+h)-(3+h)^2-(4+3(3)-(3)^2))/(h)`

`(4+9+3h-(9+6h+h^2)-(4+9-9))/(h)`

`(4+9+3h-9-6h-h^2-4)/(h)`

`(-3h-h^2)/(h)`

`(h(-3-h))/(h)`

The solution is: `-3-h = -3`