Answer
$f\left( x \right)=3{{x}^{2}}-x+2$
Work Step by Step
$f\left( x \right)=a{{x}^{2}}+bx+c$
Substitute the data point$\left( 1,4 \right)$ 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 point$\left( -1,6 \right)$ into the standard quadratic equation$f\left( x \right)=a{{x}^{2}}+bx+c$,
$6=a{{\left( -1 \right)}^{2}}+b\left( -1 \right)+c$
Substitute the data point$\left( -2,16 \right)$ into the standard quadratic equation$f\left( x \right)=a{{x}^{2}}+bx+c$,
$16=a{{\left( -2 \right)}^{2}}+b\left( -2 \right)+c$
Simplify the equations further as shown below,
$4=a+b+c$ …… (1)
$6=a-b+c$ …… (2)
$16=4a-2b+c$ …… (3)
Addition of equation (1) and (2),
$\begin{align}
& \underline{\begin{align}
& 4=a+b+c \\
& 6=a-b+c
\end{align}} \\
& \text{10}=2a+2c \\
\end{align}$
Further,
$5=a+c$ …..(4)
Multiply $\left( -2 \right)$ into the equation (2) and then add with equation (3),
$\begin{align}
& \underline{\begin{align}
& -12=-2a+2b-2c \\
& 16=4a-2b+c
\end{align}} \\
& \text{ 4}=2a-c \\
\end{align}$
Further,
$4=2a-c$ …..(5)
Add equation (4) and (5),
$\begin{align}
& \underline{\begin{align}
& 4=2a-c \\
& 5=a+c
\end{align}} \\
& \text{9}=3a \\
\end{align}$
Further,
$\begin{align}
& a=\frac{9}{3} \\
& =3
\end{align}$
Therefore,
$a=3$
Substitute $a=3$ into the equation $5=a+c$,
$\begin{align}
& 5=3+c \\
& c=5-2 \\
& =2
\end{align}$
Therefore,
$c=2$
Substitute $a=3$ and $c=2$ into the equation $4=a+b+c$,
$\begin{align}
& 4=3+b+2 \\
& 4=5+b \\
& b=-1
\end{align}$
Therefore,
$b=-1$
Put the values$a=3,b=-1$ and $c=2$ in the standard quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$,
$f\left( x \right)=3{{x}^{2}}-x+2$.
Hence, the quadratic function that fits the set of data points $\left( 1,4 \right),\left( -1,6 \right),\left( -2,16 \right)$ is $f\left( x \right)=3{{x}^{2}}-x+2$
