Answer
The difference quotient for the provided function is \[4x+2h+1\].
Work Step by Step
Consider the provided function: $f\left( x \right)=2{{x}^{2}}+x-1$.
Now, substitute $x=x+h$ in the above equation to find $f\left( x+h \right)$
That is,
$\begin{align}
& f\left( x+h \right)=2{{\left( x+h \right)}^{2}}+\left( x+h \right)-1 \\
& =2{{x}^{2}}+4xh+2{{h}^{2}}+x+h-1
\end{align}$
Now, apply the difference quotient formula,
$\begin{align}
& \frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{2{{x}^{2}}+4xh+2{{h}^{2}}+x+h-1-\left( 2{{x}^{2}}+x-1 \right)}{h} \\
& =\frac{2{{x}^{2}}+4xh+2{{h}^{2}}+x+h-1-2{{x}^{2}}-x+1}{h} \\
& =\frac{4xh+2{{h}^{2}}+h}{h} \\
& =\frac{h\left( 4x+2h+1 \right)}{h}
\end{align}$
Further solve and get,
$\frac{f\left( x+h \right)-f\left( x \right)}{h}=4x+2h+1$
Hence, the difference quotient for the provided function is $4x+2h+1$.
