Answer
The quadratic function is $y=2{{x}^{2}}-x-3$.
Work Step by Step
Let us consider the given equation:
$y=a{{x}^{2}}+bx+c$ …… (I)
If the graph of $y=a{{x}^{2}}+bx+c$ passes through the point $\left( -2,7 \right),\left( 1,-2 \right)\text{ and }\left( 2,3 \right)$ then it must satisfy the given equation.
Therefore, putting in the value of $x=-2,y=7$ in equation (I):
$7=a{{\left( -2 \right)}^{2}}+b\left( -2 \right)+c$
$7=4a-2b+c$ …… (II)
Now putting in the value of $x=1\text{ and }y=-2$:
$-2=a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+c$
$-2=a+b+c$ …… (III)
Putting the value of $x=2\text{ and }y=3$:
$3=4a+2b+c$ …… (IV)
Subtract equation (II) from equation (I):
$\begin{align}
& 4a-2b+c=7 \\
& a+b+c=-2 \\
& \underline{-\text{ }-\text{ }-\text{ + }} \\
& 3a-3b=9 \\
\end{align}$
Further simplify:
$a-b=3$ …… (V)
Now, subtract equation (III) from equation (I):
$\begin{align}
& 4a-2b+c=7 \\
& 4a+2b+c=3 \\
& \underline{-\text{ }-\text{ }-\text{ }-\text{ }} \\
& -4b=4 \\
& \text{ }b=-1 \\
\end{align}$
Putting in the value of $b=-1$ in equation (V):
$\begin{align}
& a-\left( -1 \right)=3 \\
& a=2
\end{align}$
Putting in the value of $b=-1,a=2$ in equation (II):
$\begin{align}
& 2+\left( -1 \right)+c=-2 \\
& c=-3
\end{align}$
Now, putting in the values of $a,b\text{ and }c$ in the equation $y=a{{x}^{2}}+bx+c$ then the equation is,
$y=2{{x}^{2}}-x-3$
Thus, the quadratic function is $y=2{{x}^{2}}-x-3$.
