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 $\left( -1,6 \right),\left( 1,4 \right)\text{ and }\left( 2,9 \right)$ then it must satisfy the given equation.
Therefore, putting the value of $x=-1\text{ and }y=6$ in equation (I):
$6=a{{\left( -1 \right)}^{2}}+b\left( -1 \right)+c$
$6=a-b+c$ …… (II)
Putting the value of $x=1\text{ and }y=4$:
$4=a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+c$
$4=a+b+c$ …… (III)
Now, putting the value of $x=2,y=9$
$9=4a+2b+c$ …… (IV)
Now, subtract equation (III) from equation (II):
$\begin{align}
& a-b+c=6 \\
& a+b+c=4 \\
& \underline{-\text{ }-\text{ }-\text{ }-\text{ }} \\
& -2b=2 \\
\end{align}$
Further simplify,
$b=-1$
Putting the value of $b=-1$ in equation (IV):
$\begin{align}
& 4a+2\left( -1 \right)+c=9 \\
& 4a-2+c=9 \\
\end{align}$
$4a+c=11$ …… (V)
Now, adding equation (II) and equation (III):
$\begin{align}
& a-b+c=6 \\
& \underline{a+b+c=4} \\
& 2a+2c=10 \\
\end{align}$
$a+c=5$ …… (VI)
Now, subtract the equation (VI) from equation (V):
$\begin{align}
& 4a+c=11 \\
& a+c=5 \\
& \underleftarrow{-\text{ }-\text{ }-\text{ }} \\
& 3a=6 \\
\end{align}$
Further simplify,
$a=2$
Putting in the value of $a=2$ in equation (VI):
$\begin{align}
& 2+c=5 \\
& c=3
\end{align}$
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 $y=2{{x}^{2}}-x+3$.
