How do you convert from vertex form to intercept form of $y-4=-{{\left( x-2 \right)}^{2}}$?

Hint: We use the fact that the vertex form of quadratic equation $y=a{{x}^{2}}+bx+c$ is given by ${{\left( x-h \right)}^{2}}=4a\left( y-k \right)$ where $\left( h,k \right)$ is the vertex of the parabola and the intercept form of the quadratic equation is given by $y=k\left( x-\alpha \right)\left( x-\beta \right)$ where $\alpha ,\beta $ are the $x-$intercepts of the parabola. We find the roots of the given quadratic equation to find the intercept form. \[\]

Complete step-by-step answer:
We know that the general quadratic equation is given by $a{{x}^{2}}+bx+c=0$ where graph of the quadratic polynomial $y=a{{x}^{2}}+bx+c$ represents a parabola in $xy-$plane. The vertex from of the parabola is given by with vertex $\left( h,k \right)$ is given by
\[{{\left( x-h \right)}^{2}}=4a\left( y-k \right)\]
We are given the following quadratic equation in the question
\[y-4=-{{\left( x-2 \right)}^{2}}\]
We expand the right hand side using the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$to have
\[\begin{align}
  & \Rightarrow y-4=-\left( {{x}^{2}}-8x+16 \right) \\
 & \Rightarrow y=-{{x}^{2}}+8x-12 \\
\end{align}\]
So we need to convert the above result to the intercept form $y=k\left( x-\alpha \right)\left( x-\beta \right)$. So we have to find the roots since we have $x-$intercepts at $y=0$. So let us have
\[\begin{align}
  & y=0 \\
 & \Rightarrow -{{x}^{2}}+8x-12=0 \\
 & \Rightarrow {{x}^{2}}-8x-12=0 \\
\end{align}\]
We use the splitting the middle term method to have;
\[\begin{align}
  & \Rightarrow {{x}^{2}}-6x-2x+12=0 \\
 & \Rightarrow x\left( x-6 \right)-2\left( x-6 \right)=0 \\
 & \Rightarrow \left( x-6 \right)\left( x-2 \right)=0 \\
 & \Rightarrow x=6,2 \\
\end{align}\]
So we have the roots of the quadratic polynomial as $\alpha =6,\beta =2$. Since we have multiplied $-1$ both sides and then solved the equation we have the required intercept form is
\[y=-\left( x-2 \right)\left( x-6 \right)\]

Note: We note that the vertex of the given equation $y-4=-{{\left( x-2 \right)}^{2}}$ is at $\left( 2,4 \right)$ and also note that at the vertex the minimum values of the quadratic polynomial occurs. So at $x=4$ we are getting the minimum as $y=-\left( 4-2 \right)\left( 4-6 \right)=4$. So our equation in intercept form is verified. We can find the vertex from general quadratic polynomial $y=a{{x}^{2}}+bx+c$ at $x=\dfrac{-b}{2a}$ . If in $y=a{{x}^{2}}+bx+c$ we have $a > 0$ we get a parabola opens upwards and if we have $a < 0$ we get a parabola opens downwards.