Answer
The quadratic equation is: $-\frac{3}{4}{{x}^{2}}+6x-11$.
Work Step by Step
The given quadratic equation is as shown below:
$y=a{{x}^{2}}+bx+c$
The given points are $\left( 4,1 \right),\left( 2,-2 \right),\left( 6,-2 \right)$. To find the values of a, b, and c of the quadratic equation, substitute the following pairs in the equation.
Substitute $x=4$ and $y=1$
$\begin{align}
& 1=a{{\left( 4 \right)}^{2}}+b\left( 4 \right)+c \\
& 1=16a+4b+c
\end{align}$
Substitute $x=2$ and $y=-2$
$\begin{align}
& -2=a{{\left( 2 \right)}^{2}}+b\left( 2 \right)+c \\
& -2=4a+2b+c
\end{align}$
Substitute $x=6$ and $y=-2$
$\begin{align}
& -2=a{{\left( 6 \right)}^{2}}+b\left( 6 \right)+c \\
& -2=36a+6b+c
\end{align}$
Mark the resultant system of equations as shown below:
$16a+4b+c=1$ (I)
$4a+2b+c=-2$ (II)
$36a+6b+c=-2$ (III)
By multiplying equation (III) by -1 and eliminating c from equations (I) and (III), we get
$\begin{align}
& \text{ }16a+4b+c=\text{ }1 \\
& \underline{-36a-6b-c\text{ }=\text{ 2}} \\
& \text{ }-\text{20}a-2b\text{ }=\text{ }3 \\
\end{align}$ (IV)
By multiplying the equation (III) by -1 and eliminating c from equations (II) and (III), we get
$\begin{align}
& \text{ }4a+2b+\text{ }c=-2 \\
& \underline{-36a-6b-c=\text{ 2}} \\
& -32a-4b\text{ }=\text{ }0 \\
\end{align}$ (V)
By multiplying the equation (IV) by -2 and then adding equations (IV) and (V), we get the value of a,
$\begin{align}
& -32a-4b=\text{ }0 \\
& \underline{\text{ }40a+4b=-6} \\
& \text{ }8a=-6 \\
\end{align}$
Now, divide both sides by 8:
$\begin{align}
& \frac{8a}{8}=\frac{-6}{8} \\
& \text{ }a=-\frac{3}{4} \\
\end{align}$
Substitute the value of a in equation (IV) to get:
$\begin{align}
& -\text{20}\left( -\frac{3}{4} \right)-2b=3\text{ } \\
& 15-2b=3
\end{align}$
Now, subtract 15 from both sides,
$\begin{align}
& 15-2b-15=3-15 \\
& -2b=-12
\end{align}$
By dividing both sides by -2, we get
$\begin{align}
& \frac{-2b}{-2}=\frac{-12}{-2} \\
& \text{ }b=6 \\
\end{align}$
Substitute the values of b and a in equation (II) to get the value of c,
$\begin{align}
& 4\left( -\frac{3}{4} \right)+2\left( 6 \right)+c=-2 \\
& -3+12+c=-2 \\
& 9+c=-2
\end{align}$
Now, subtract $9$ from both sides and we get
$\begin{align}
& 9+c-9=-2-9 \\
& \text{ }c=-11 \\
\end{align}$
Thus, the quadratic equation is $-\frac{3}{4}{{x}^{2}}+6x-11$.
