Answer
The quadratic equation is $2{{x}^{2}}+x-5$.
Work Step by Step
The standard quadratic equation is shown below:
$y=a{{x}^{2}}+bx+c$
The points are $\left( -1,-4 \right),\left( 1,-2 \right),\left( 2,5 \right)$. In order to get the quadratic function, find the values of a, b, and c, and substitute the values of the ordered pairs in the quadratic function.
Put $x=-1$ and $y=-4$
$\begin{align}
& -4=a{{\left( -1 \right)}^{2}}+b\left( -1 \right)+c \\
& -4=a-b+c
\end{align}$ (I)
Put $x=1$ and $y=-2$
$\begin{align}
& -2=a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+c \\
& -2=a+b+c
\end{align}$ (II)
Put $x=2$ and $y=5$
$\begin{align}
& 5=a{{\left( 2 \right)}^{2}}+b\left( 2 \right)+c \\
& 5=4a+2b+c
\end{align}$ (III)
Get the system of equations as follows:
$\begin{align}
& a-b+c=-4 \\
& a+b+c=-2 \\
& 4a+2b+c=5
\end{align}$
Eliminate a from equations (I) and (II) to get;
$\begin{align}
& \text{ }a-b+c=-4 \\
& -a-b-c=2 \\
\end{align}$
$-2b=-2$
Divide both sides by $-2$ to get:
$\begin{align}
& \frac{-2b}{-2}=\frac{-2}{-2} \\
& b=1
\end{align}$
Putting in the value of b in equation (I): we get
$\begin{align}
& a-\left( 1 \right)+c=-4 \\
& a+c-1=-4
\end{align}$
Now, adding 1 to both sides to get:
$\begin{align}
& a+c-1+1=-4+1 \\
& a+c=-3
\end{align}$ (IV)
Putting the value of b in equation (III) to get:
$\begin{align}
& 4a+2\left( 1 \right)+c=5 \\
& 4a+2+c=5
\end{align}$
Subtract 2 from both sides to get,
$\begin{align}
& 4a+c+2-2=5-2 \\
& 4a+c=3
\end{align}$ (V)
By eliminating c from equations (IV) and (V)
$\begin{align}
& 4a+c=\text{3} \\
& -a-c=\text{3}
\end{align}$
$3a\text{ }=\text{ }6$ (VI)
By dividing both sides by 3: we get,
$\begin{align}
& \frac{3a}{3}=\frac{6}{3} \\
& a=2
\end{align}$
Putting the value of a in equation (IV) to find the value of c: we get
$2+c=-3$
Now, subtract 2 from both sides to get,
$\begin{align}
& 2+c-2=-3-2 \\
& c=-5
\end{align}$
Now putting the values of a, b and c to get the equation $2{{x}^{2}}+x-5$.
Hence, the quadratic equation is $2{{x}^{2}}+x-5$.
