How do you find the polynomial function whose graph passes through $\left( 2,4 \right),\left( 3,6 \right),\left( 5,10 \right)?$

Hint: We will use second polynomial fitting. Then we will find the corresponding coefficient matrix. Later, we will use the Cramer’s Rule to determine the unknowns.

Complete step by step solution:
We are given with three points, $\left( 2,4 \right),\left( 3,6 \right)$ and $\left( 5,10 \right).$
Therefore, we will use second polynomial fitting like $y=a{{x}^{2}}+bx+c.$
We will get the following three equations:
When we consider the point $\left( 2,4 \right),$ we will get ${{y}_{1}}=ax_{1}^{2}+b{{x}_{1}}+c=a{{\left( 2 \right)}^{2}}+b\left( 2 \right)+c.$
When we consider the point $\left( 3,6 \right),$ we will get ${{y}_{2}}=ax_{2}^{2}+b{{x}_{2}}+c=a{{\left( 3 \right)}^{2}}+b\left( 3 \right)+c.$
When we consider the point $\left( 5,10 \right),$ we will get ${{y}_{3}}=ax_{3}^{2}+b{{x}_{3}}+c=a{{\left( 5 \right)}^{2}}+b\left( 5 \right)+c.$
These equations will give us the following system of equations:
$\begin{align}
  & 4a+2b+c=4. \\
 & 9a+3b+c=6. \\
 & 25a+5b+c=10. \\
\end{align}$
Let us convert this system of equations into matrix form as $AX=B$ where $A=\left[ \begin{align}
  & \begin{matrix}
   4 & 2 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   9 & 3 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   25 & 5 & 1 \\
\end{matrix} \\
\end{align} \right], X=\left[ \begin{matrix}
   a \\
   b \\
   c \\
\end{matrix} \right]$ and $B=\left[ \begin{matrix}
   4 \\
   6 \\
   10 \\
\end{matrix} \right].$
When we solve this using $X={{A}^{-1}}B,$ we will get the values the unknowns $a,b$ and $c.$
Here we are using Cremer’s rule to find the values of the unknowns $a,b$ and $c.$
Let us find the determinant of the matrix $A.$
$\Rightarrow \left| A \right|=\left| \begin{align}
  & \begin{matrix}
   4 & 2 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   9 & 3 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   25 & 5 & 1 \\
\end{matrix} \\
\end{align} \right|=4\left( 3-5 \right)-2\left( 9-25 \right)+1\left( 45-75 \right)=4\times \left( -2 \right)-2\left( -16 \right)+\left( -30 \right)$
$\Rightarrow \left| A \right|=-8+32-30=-8+2=-6.$
Consider the matrix ${{A}_{{{x}_{1}}}}$ which is constructed by replacing the first column of the matrix $A$ with the elements of the column matrix $B.$
So, the matrix ${{A}_{{{x}_{1}}}}$ is given by ${{A}_{{{x}_{1}}}}=\left[ \begin{align}
  & \begin{matrix}
   4 & 2 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   6 & 3 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   10 & 5 & 1 \\
\end{matrix} \\
\end{align} \right].$
We are going to find the determinant of the matrix ${{A}_{{{x}_{1}}}}.$
Therefore, $\left| {{A}_{{{x}_{1}}}} \right|=4\left( 3-5 \right)-2\left( 6-10 \right)+1\left( 30-30 \right)=4\times \left( -2 \right)-2\left( -4 \right)+0=-8+8=0.$
Consider the matrix ${{A}_{{{x}_{2}}}}$ which is constructed by replacing the second column of the matrix $A$ with the elements of the column matrix $B.$
So, the matrix \[{{A}_{{{x}_{2}}}}=\left[ \begin{align}
  & \begin{matrix}
   4 & 4 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   9 & 6 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   25 & 10 & 1 \\
\end{matrix} \\
\end{align} \right].\]
As we have done earlier, we are going to find the determinant of the matrix ${{A}_{y}}.$
So, \[\left| {{A}_{{{x}_{2}}}} \right|=4\left( 6-10 \right)-4\left( 9-25 \right)+1\left( 90-150 \right)=4\left( -4 \right)-4\left( -16 \right)+\left( -60 \right)=-16+64-60=-16+4=-12.\]
Consider the matrix ${{A}_{{{x}_{3}}}}$ which is constructed by replacing the third column of the matrix $A$ with the elements of the column matrix $B.$
Thus, the matrix ${{A}_{{{x}_{3}}=}}\left[ \begin{align}
  & \begin{matrix}
   4 & 2 & 4 \\
\end{matrix} \\
 & \begin{matrix}
   9 & 3 & 6 \\
\end{matrix} \\
 & \begin{matrix}
   25 & 5 & 10 \\
\end{matrix} \\
\end{align} \right].$
The determinant of ${{A}_{{{x}_{3}}}}=4\left( 30-30 \right)-2\left( 90-150 \right)+4\left( 45-75 \right)=0-2\left( -60 \right)+4\left( -30 \right)=120-120=0.$
Now, the values of the unknowns can be obtained as follows:
$a=\dfrac{\left| {{A}_{{{x}_{1}}}} \right|}{\left| A \right|}=\dfrac{0}{-6}=0.$
$b=\dfrac{\left| {{A}_{{{x}_{2}}}} \right|}{\left| A \right|}=\dfrac{-12}{-6}=\dfrac{12}{6}=2.$
$c=\dfrac{\left| {{A}_{{{x}_{3}}}} \right|}{\left| A \right|}=\dfrac{0}{-6}=0.$
Now we will substitute these values in any of the equations in the system of equations.
We will get,
$\begin{align}
  & 4\left( 0 \right)+2\left( 2 \right)+0=4. \\
 & 9\left( 0 \right)+3\left( 2 \right)+0=6. \\
 & 25\left( 0 \right)+5\left( 2 \right)+0=10. \\
\end{align}$
We can see,
$\begin{align}
  & {{y}_{1}}=0x_{1}^{2}+2{{x}_{1}}+0=2{{x}_{1}}. \\
 & {{y}_{2}}=0x_{2}^{2}+2{{x}_{2}}+0=2{{x}_{2}}. \\
 & {{y}_{3}}=0x_{3}^{2}+2{{x}_{3}}+0=2{{x}_{3}}. \\
\end{align}$
Generally, $y=2x.$

Hence the polynomial function whose graph passes through $\left( 2,4 \right),\left( 3,6 \right)$ and $\left( 5,10 \right)$ $y=2x.$

Note: In $\left( 2,4 \right), x=2$ and $y=4.$ That is $y=2\cdot 2=2x.$ In $\left( 3,6 \right), x=3$ and $y=6.$ That is, $y=2\cdot 3=2x.$ Similarly, in $\left( 5,10 \right), x=5$ and $y=10.$ That is, $y=2\cdot 5=2x.$ Therefore the required function is $y=2x.$
The determinant $\left| \begin{align}
  & \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
\end{matrix} \\
 & \begin{matrix}
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
\end{matrix} \\
 & \begin{matrix}
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \\
\end{align} \right|={{a}_{1}}\left( {{b}_{2}}{{c}_{3}}-{{c}_{2}}{{b}_{3}} \right)-{{a}_{2}}\left( {{b}_{1}}{{c}_{3}}-{{c}_{1}}{{b}_{3}} \right)+{{a}_{3}}\left( {{b}_{1}}{{c}_{2}}-{{c}_{1}}{{b}_{2}} \right).$