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}\over{2}}#, #b=5# 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:
\[\begin{array}{rclrl}
x&=& -\dfrac{5}{2 \cdot -{{1}\over{2}}} &&\phantom{xxx}\blue{\text{formula entered}}\\
&=& 5 &&\phantom{xxx}\blue{\text{simplified}}\\
\end{array}\]
The #y#-value of the vertex is calculated by entering #x=5# in the formula. Which gives:
\[\begin{array}{rclrl}
y&=& -{{1}\over{2}} \cdot 5^2 +5 \cdot 5 +6
&&\phantom{xxx}\blue{\text{formula entered}}\\
&=& {{37}\over{2}} &&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the vertex are: #\rv{5,{{37}\over{2}}}#. To draw the point in the graph, we have to write the coordinates as decimal numbers (rounded to 1 decimal). That gives: #\rv{5,18.5}#.

The intersections with the #x#-axis are the points that correspond to #y=0#.
\[\begin{array}{rcl}
-{{x^2}\over{2}}+5\cdot x+6 &=& 0 \\&&\phantom{xxx}\blue{\text{the equation that should be calculated}}\\
x=\dfrac{-{5}-\sqrt{5^2-4 \cdot -{{1}\over{2}} \cdot 6}}{2 \cdot -{{1}\over{2}}} &\vee& x=\dfrac{-{5}+\sqrt{5^2-4 \cdot -{{1}\over{2}} \cdot 6}}{2 \cdot -{{1}\over{2}}} \\&&\phantom{xxx}\blue{\text{quadratic formula entered}}\\
x=5-\sqrt{37} &\vee& x=\sqrt{37}+5 \\&&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the intersections with the #x#-axis are: #\rv{5-\sqrt{37},0}# and #\rv{\sqrt{37}+5,0}#. To draw the point in the graph, we have to write the coordinates as decimal numbers (rounded to 1 decimal). That gives:#\rv{-1.1,0}# and #\rv{11.1,0}#.

The four points in the graph are: #\rv{0,6}#, #\rv{5,{{37}\over{2}}}#, #\rv{5-\sqrt{37},0}# and #\rv{\sqrt{37}+5,0}#.