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=2# and #c=-5#. Seeing as #a>0# the graph is a parabola that opens upward.

The intersection with the #y#-axis is equal to the value of the constant in the quadratic formula, which is equal to #-5#. That means that the coordinates of the intersection point with the #y#-axis are #\rv{0,-5}#.

The #x#-value of the vertex is given by #x=-\dfrac{b}{2 \cdot a}# and is equal to:
\[\begin{array}{rclrl}
x&=& -\dfrac{2}{2 \cdot {{1}\over{2}}} &&\phantom{xxx}\blue{\text{formula entered}}\\
&=& -2 &&\phantom{xxx}\blue{\text{simplified}}\\
\end{array}\]
The #y#-value of the vertex is calculated by entering #x=-2# in the formula. Which gives:
\[\begin{array}{rclrl}
y&=& {{1}\over{2}} \cdot \left(-2\right)^2 +2 \cdot -2 -5
&&\phantom{xxx}\blue{\text{formula entered}}\\
&=& -7 &&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the vertex are: #\rv{-2,-7}#.

The intersections with the #x#-axis are the points that correspond to #y=0#.
\[\begin{array}{rcl}
{{x^2}\over{2}}+2\cdot x-5 &=& 0 \\&&\phantom{xxx}\blue{\text{the equation that should be calculated}}\\
x=\dfrac{-{2}-\sqrt{2^2-4 \cdot {{1}\over{2}} \cdot -5}}{2 \cdot {{1}\over{2}}} &\vee& x=\dfrac{-{2}+\sqrt{2^2-4 \cdot {{1}\over{2}} \cdot -5}}{2 \cdot {{1}\over{2}}} \\&&\phantom{xxx}\blue{\text{quadratic formula entered}}\\
x=-\sqrt{14}-2 &\vee& x=\sqrt{14}-2 \\&&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the intersections with the #x#-axis are: #\rv{-\sqrt{14}-2,0}# and #\rv{\sqrt{14}-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{-5.7,0}# and #\rv{1.7,0}#.

The four points in the graph are: #\rv{0,-5}#, #\rv{-2,-7}#, #\rv{-\sqrt{14}-2,0}# and #\rv{\sqrt{14}-2,0}#.