How do you write a quadratic equation with vertex $\left( 2,-4 \right)$ and the point $\left( 1,-1 \right)$?

Hint: We have to find the quadratic equation with vertex $\left( 2,-4 \right)$ and an arbitrary point on which the curve goes through is $\left( 1,-1 \right)$. The equation will be a general equation of parabola. We equate the required equation of parabolic curve with the general equation of ${{\left( x-\alpha \right)}^{2}}=4a\left( y-\beta \right)$. We put the values of the arbitrary point and the vertex coordinate.

Complete step-by-step solution:
The general equation ${{\left( x-\alpha \right)}^{2}}=4a\left( y-\beta \right)$ is a parabolic curve.
For the general equation $\left( \alpha ,\beta \right)$ is the vertex. 4a is the length of the latus rectum. The coordinate of the focus is $\left( \alpha ,\beta +a \right)$.
This gives the vertex as $\left( 2,-4 \right)$. Putting the value in ${{\left( x-\alpha \right)}^{2}}=4a\left( y-\beta \right)$, we get
$\begin{align}
  & {{\left( x-\alpha \right)}^{2}}=4a\left( y-\beta \right) \\
 & \Rightarrow {{\left( x-2 \right)}^{2}}=4a\left[ y-\left( -4 \right) \right] \\
 & \Rightarrow {{\left( x-2 \right)}^{2}}=4a\left[ y+4 \right] \\
\end{align}$
We need to find the value of $a$.
We now have the arbitrary point $\left( 1,-1 \right)$. The curve ${{\left( x-2 \right)}^{2}}=4a\left[ y+4 \right]$ goes through the point $\left( 1,-1 \right)$. Putting the value of $\left( 1,-1 \right)$ in the equation, we get
\[\begin{align}
  & {{\left( 1-2 \right)}^{2}}=4a\left[ -1+4 \right] \\
 & \Rightarrow {{\left( -1 \right)}^{2}}=4a\times 3 \\
 & \Rightarrow 12a=1 \\
 & \Rightarrow 4a=\dfrac{1}{3} \\
\end{align}\]
Putting value of $4a$ in the equation ${{\left( x-2 \right)}^{2}}=4a\left( y+4 \right)$, we get
$\begin{align}
  & {{\left( x-2 \right)}^{2}}=\dfrac{1}{3}\left( y+4 \right) \\
 & \Rightarrow 3{{\left( x-2 \right)}^{2}}=\left( y+4 \right) \\
\end{align}$
Now simplifying we get
$\begin{align}
  & 3{{\left( x-2 \right)}^{2}}=\left( y+4 \right) \\
 & \Rightarrow 3{{x}^{2}}-12x+12-4=y \\
 & \Rightarrow y=3{{x}^{2}}-12x+8 \\
\end{align}$
The quadratic equation is $y=3{{x}^{2}}-12x+8$.


Note: The minimum point of the function $y=3{{x}^{2}}-12x+8$ is $\left( 2,-4 \right)$. The graph is bounded at that point. But on the other side the curve is open and not bounded. The general case of parabolic curve is to be bounded at one side to mark the vertex.