Answer
$f\left( x \right)=2{{x}^{2}}+3x-1$.
Work Step by Step
$f\left( x \right)=a{{x}^{2}}+bx+c$
Substitute the data points $\left( 1,4 \right)$ into the standard quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$,
$4=a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+c$
Substitute the data points $\left( -1,-2 \right)$ into the standard quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$,
$-2=a{{\left( -1 \right)}^{2}}-b\left( -1 \right)+c$
Substitute the data points $\left( 2,13 \right)$into the standard quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$,
$13=a{{\left( 2 \right)}^{2}}+b\left( 2 \right)+c$
Simplify the equations further as shown below,
$4=a+b+c$ …… (1)
$-2=a-b+c$ …… (2)
$13=4a+2b+c$ …… (3)
Thus:
$\begin{align}
& \underline{\begin{align}
& 4=a+b+c \\
& -2=a-b+c
\end{align}} \\
& \text{ }2=2a+2c \\
\end{align}$
Further,
$1=a+c$ …… (4)
Multiply 2 into the equation (2) and then add with equation (3),
$\begin{align}
& \underline{\begin{align}
& -4=2a-2b+2c \\
& 13=4a+2b+c
\end{align}} \\
& \text{ 9}=6a+3c \\
\end{align}$
Further,
$\text{3}=2a+c$ …… (5)
Subtract equation (4) by equation (5),
$\begin{align}
& \underline{\begin{align}
& \text{3}=2a+c \\
& -1=-a-c
\end{align}} \\
& \text{ 2}=a \\
\end{align}$
Thus,
$a=2$
Substitute $a=2$ into the equation $1=a+c$,
$\begin{align}
& 1=2+c \\
& c=1-2 \\
& =-1
\end{align}$
Therefore,
$c=-1$
Substitute $a=2$ and $c=-1$ into the equation $4=a+b+c$,
$\begin{align}
& 4=2+b+\left( -1 \right) \\
& 4=1+b \\
& b=3
\end{align}$
Thus,
$b=3$
Put the values $a=2,b=3$ and $c=-1$ in the standard quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$,
$f\left( x \right)=2{{x}^{2}}+3x-1$
Hence, the quadratic function that fits the set of data points $\left( 1,4 \right),\left( -1,-2 \right),\left( 2,13 \right)$ is $f\left( x \right)=2{{x}^{2}}+3x-1$.
