The red dots are the four dots from the question. These are calculated as follows:
The formula is already written in the form of #a \cdot x^2+b \cdot x +c# with #a =-1#, #b=4# and #c=6#. Seeing as #a<0# the graph is a parabola that opens downward.

The intersection with the #y#-axis is equal to the value of the constant in the quadratic formula, which is equal to #6#. That means that the coordinates of the intersection point with the #y#-axis are #\rv{0,6}#.

The #x#-value of the vertex is given by #x=-\dfrac{b}{2 \cdot a}# and is equal to:
\[\begin{array}{rclrl}
x&=& -\dfrac{4}{2 \cdot -1} &&\phantom{xxx}\blue{\text{formula entered}}\\
&=& 2 &&\phantom{xxx}\blue{\text{simplified}}\\
\end{array}\]
The #y#-value of the vertex is calculated by entering #x=2# in the formula. Which gives:
\[\begin{array}{rclrl}
y&=& -2^2 +4 \cdot 2 +6
&&\phantom{xxx}\blue{\text{formula entered}}\\
&=& 10 &&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the vertex are: #\rv{2,10}#.

The intersections with the #x#-axis are the points that correspond to #y=0#.
\[\begin{array}{rcl}
-x^2+4\cdot x+6 &=& 0 \\&&\phantom{xxx}\blue{\text{the equation that should be calculated}}\\
x=\dfrac{-{4}-\sqrt{4^2-4 \cdot -1 \cdot 6}}{2 \cdot -1} &\vee& x=\dfrac{-{4}+\sqrt{4^2-4 \cdot -1 \cdot 6}}{2 \cdot -1} \\&&\phantom{xxx}\blue{\text{quadratic formula entered}}\\
x=2-\sqrt{10} &\vee& x=\sqrt{10}+2 \\&&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the intersections with the #x#-axis are: #\rv{2-\sqrt{10},0}# and #\rv{\sqrt{10}+2,0}#. To draw the point in the graph, we have to write the coordinates as decimal numbers (rounded to 1 decimal). That gives:#\rv{-1.2,0}# and #\rv{5.2,0}#.

The four points in the graph are: #\rv{0,6}#, #\rv{2,10}#, #\rv{2-\sqrt{10},0}# and #\rv{\sqrt{10}+2,0}#.