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=-2# and #c=-2#. 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 #-2#. That means that the coordinates of the intersection point with the #y#-axis are #\rv{0,-2}#.

The #x#-value of the vertex is given by #x=-\dfrac{b}{2 \cdot a}# and is equal to:
\[\begin{array}{rclrl}
x&=& -\dfrac{-2}{2 \cdot 1} &&\phantom{xxx}\blue{\text{formula entered}}\\
&=& 1 &&\phantom{xxx}\blue{\text{simplified}}\\
\end{array}\]
The #y#-value of the vertex is calculated by entering #x=1# in the formula. Which gives:
\[\begin{array}{rclrl}
y&=& 1^2 -2 \cdot 1 -2
&&\phantom{xxx}\blue{\text{formula entered}}\\
&=& -3 &&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the vertex are: #\rv{1,-3}#.

The intersections with the #x#-axis are the points that correspond to #y=0#.
\[\begin{array}{rcl}
x^2-2\cdot x-2 &=& 0 \\&&\phantom{xxx}\blue{\text{the equation that should be calculated}}\\
x=\dfrac{-{-2}-\sqrt{\left(-2\right)^2-4 \cdot 1 \cdot -2}}{2 \cdot 1} &\vee& x=\dfrac{-{-2}+\sqrt{\left(-2\right)^2-4 \cdot 1 \cdot -2}}{2 \cdot 1} \\&&\phantom{xxx}\blue{\text{quadratic formula entered}}\\
x=1-\sqrt{3} &\vee& x=\sqrt{3}+1 \\&&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the intersections with the #x#-axis are: #\rv{1-\sqrt{3},0}# and #\rv{\sqrt{3}+1,0}#. To draw the point in the graph, we have to write the coordinates as decimal numbers (rounded to 1 decimal). That gives:#\rv{-0.7,0}# and #\rv{2.7,0}#.

The four points in the graph are: #\rv{0,-2}#, #\rv{1,-3}#, #\rv{1-\sqrt{3},0}# and #\rv{\sqrt{3}+1,0}#.