Answer
$f\left( x \right)=-\frac{1}{4}{{x}^{2}}+3x-5$
Work Step by Step
$f\left( x \right)=a{{x}^{2}}+bx+c$
Substitute the data point$\left( 2,0 \right)$ into the standard quadratic equation$f\left( x \right)=a{{x}^{2}}+bx+c$,
$0=a{{\left( 2 \right)}^{2}}+b\left( 2 \right)+c$
Substitute the data point$\left( 4,3 \right)$ into the standard quadratic equation$f\left( x \right)=a{{x}^{2}}+bx+c$,
$3=a{{\left( 4 \right)}^{2}}+b\left( 4 \right)+c$
Substitute the data point$\left( 12,-5 \right)$ into the standard quadratic equation$f\left( x \right)=a{{x}^{2}}+bx+c$,
$-5=a{{\left( 12 \right)}^{2}}+b\left( 12 \right)+c$
Simplify the equations further as shown below,
$0=4a+2b+c$ …… (1)
$3=16a+4b+c$ …… (2)
$-5=144a+12b+c$ …… (3)
Subtract equation (2) from (1),
$\begin{align}
& \underline{\begin{align}
& 0=4a+2b+c \\
& -3=-16a-4b-c
\end{align}} \\
& -3=-12a-2b \\
\end{align}$
Further,
$-3=-12a-2b$ …...(4)
Subtract equation (2) from (3),
$\begin{align}
& \underline{\begin{align}
& -5=144a+12b+c \\
& -3=-16a-4b-c
\end{align}} \\
& -8=128a+8b \\
\end{align}$
Simplify further,
$-2=32a+2b$ …..(5)
Add equation (4) and (5),
$\begin{align}
& \underline{\begin{align}
& -2=32a+2b \\
& -3=-12a-2b
\end{align}} \\
& -5=20a \\
\end{align}$
Further,
$\begin{align}
& a=-\frac{20}{5} \\
& =-\frac{1}{4}
\end{align}$
Therefore,
$a=-\frac{1}{4}$
Substitute $a=-\frac{1}{4}$ into the equation $-3=-12a-2b$,
$\begin{align}
& -3=-12\left( -\frac{1}{4} \right)-2b \\
& 3=3-2b \\
& -6=-2b \\
& b=3
\end{align}$
Therefore,
$b=3$
Substitute $a=-\frac{1}{4}$ and $b=3$ into the equation $0=4a+2b+c$,
$\begin{align}
& 0=4\left( -\frac{1}{4} \right)+2\left( 3 \right)+c \\
& 0=1+6+c \\
& c=-5
\end{align}$
Therefore,
$c=-5$
Substitute the values $a=-\frac{1}{4},b=3$ and $c=-5$ in the standard quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$
$f\left( x \right)=-\frac{1}{4}{{x}^{2}}+3x-5$.
Hence, the quadratic function that fits the set of data points $\left( 2,0 \right),\left( 4,3 \right),\left( 12,-5 \right)$ is $f\left( x \right)=-\frac{1}{4}{{x}^{2}}+3x-5$
