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=-7#. 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 #-7#. That means that the coordinates of the intersection point with the #y#-axis are #\rv{0,-7}#.

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:
\[\begin{array}{rclrl}
y&=& 2^2 -4 \cdot 2 -7
&&\phantom{xxx}\blue{\text{formula entered}}\\
&=& -11 &&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the vertex are: #\rv{2,-11}#.

The intersections with the #x#-axis are the points that correspond to #y=0#.
\[\begin{array}{rcl}
x^2-4\cdot x-7 &=& 0 \\&&\phantom{xxx}\blue{\text{the equation that should be calculated}}\\
x=\dfrac{-{-4}-\sqrt{\left(-4\right)^2-4 \cdot 1 \cdot -7}}{2 \cdot 1} &\vee& x=\dfrac{-{-4}+\sqrt{\left(-4\right)^2-4 \cdot 1 \cdot -7}}{2 \cdot 1} \\&&\phantom{xxx}\blue{\text{quadratic formula entered}}\\
x=2-\sqrt{11} &\vee& x=\sqrt{11}+2 \\&&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the intersections with the #x#-axis are: #\rv{2-\sqrt{11},0}# and #\rv{\sqrt{11}+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{-1.3,0}# and #\rv{5.3,0}#.

The four points in the graph are: #\rv{0,-7}#, #\rv{2,-11}#, #\rv{2-\sqrt{11},0}# and #\rv{\sqrt{11}+2,0}#.