#### Answer

Using $f\left( x \right)={{x}^{2}}-3x+5$, we can determine that ${f}'\left( x \right)=2x-3$. This means that the point-slope equation of the tangent line to the graph of $f\left( x \right)={{x}^{2}}-3x+5$ at $\left( 6,23 \right)$ is $y-23=9\left( x-6 \right)$.

#### Work Step by Step

Consider the function $f\left( x \right)={{x}^{2}}-3x+5$
Since it is given that ${f}'\left( x \right)=2x-3$ ,
Find the slope of the tangent line to the graph of $f\left( x \right)={{x}^{2}}-3x+5$ at $\left( 6,23 \right)$, by substituting
$x=6$ in ${f}'\left( x \right)=2x-3$ ,
${f}'\left( 6 \right)=2\left( 6 \right)-3=9$
Thus, the slope of the tangent line to the graph of $f\left( x \right)={{x}^{2}}-3x+5$ at $\left( 6,23 \right)$ is $9$.
The point-slope equation of the tangent line to the graph of $f\left( x \right)={{x}^{2}}-3x+5$ at $\left( 6,23 \right)$ with the slope $9$ is:
Substitute the value of ${{x}_{1}}=6,{{y}_{1}}=23\text{ and }m=9$ in $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$
$y-23=9\left( x-6 \right)$
Therefore, the tangent line to the graph of $f\left( x \right)={{x}^{2}}-3x+5$ at $\left( 6,23 \right)$ is $y-23=9\left( x-6 \right)$.