Solution:
Given:
The difference quotient;
[tex]\frac{f(x+h)-f(x)}{h}[/tex]
where;
[tex]f(x)=x^2-2x+3[/tex][tex]\begin{gathered} \text{If} \\ f(x)=x^2-2x+3,\text{ then} \\ f(x+h)=(x+h)^2-2(x+h)+3 \end{gathered}[/tex]
Substituting the two functions above in the difference quotient formula;
[tex]\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{(x+h)^2-2(x+h)+3-(x^2-2x+3)}{h} \\ =\frac{(x+h)(x+h)-2x-2h+3-x^2+2x-3^{}}{h} \\ =\frac{(x^2+2xh+h^2)-2x-2h+3-x^2+2x-3^{}}{h} \\ \text{Collecting the like terms,} \\ =\frac{x^2-x^2-2x+2x+3-3+2xh+h^2-2h^{}}{h} \\ =\frac{2xh+h^2-2h}{h} \\ \text{Factorizing h from the numerator;} \\ =\frac{h(2x+h-2)}{h} \\ \text{Cancelling out the h in the numerator and denominator, we have;} \\ 2x+h-2 \\ =2x-2+h \end{gathered}[/tex]
Therefore, the difference quotient is;
[tex]2x-2+h[/tex]
