Answer
The difference quotient for the provided function is $2x+h-4$.
Work Step by Step
Consider the provided function: $f\left( x \right)={{x}^{2}}-4x+3$.
Now, substitute $x=x+h$ in the above equation to find $f\left( x+h \right)$
That is,
$\begin{align}
& f\left( x+h \right)={{\left( x+h \right)}^{2}}-4\left( x+h \right)+3 \\
& ={{x}^{2}}+2xh+{{h}^{2}}-4x-4h+3
\end{align}$
Now, apply the difference quotient formula,
$\begin{align}
& \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{{{x}^{2}}+2xh+{{h}^{2}}-4x-4h+3-\left( {{x}^{2}}-4x+3 \right)}{h} \\
& =\frac{{{x}^{2}}+2xh+{{h}^{2}}-4x-4h+3-{{x}^{2}}+4x-3}{h} \\
& =\frac{h\left( 2x+h-4 \right)}{h} \\
& =2x+h-4
\end{align}$
Hence, the difference quotient for the provided function is $2x+h-4$.
