The red dots are the four dots from the question. These are calculated as followed:
The formula is already written in the form of #a \cdot x^2+b \cdot x +c# with #a ={{1}\over{4}}#, #b=-1# 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:
\[\begin{array}{rclrl}
x&=& -\dfrac{-1}{2 \cdot {{1}\over{4}}} &&\phantom{xxx}\color{blue}{\text{formula entered}}\\
&=& 2 &&\phantom{xxx}\color{blue}{\text{simplified}}\\
\end{array}\]
The #y#-value of the vertex is calculated by entering #x=2# in the formula. Which gives:
\[\begin{array}{rclrl}
y&=& {{1}\over{4}} \cdot 2^2 -2 -5
&&\phantom{xxx}\color{blue}{\text{formula entered}}\\
&=& -6 &&\phantom{xxx}\color{blue}{\text{calculated}}\\
\end{array}\]
The coordinates of the vertex are: #\rv{2,-6}#.

The intersections with the #x#-axis are the points that correspond to #y=0#.
\[\begin{array}{rcl}
{{x^2}\over{4}}-x-5 &=& 0 \\&&\phantom{xxx}\color{blue}{\text{the equation that should be calculated}}\\
x=\dfrac{-{-1}-\sqrt{\left(-1\right)^2-4 \cdot {{1}\over{4}} \cdot -5}}{2 \cdot {{1}\over{4}}} &\vee& x=\dfrac{-{-1}+\sqrt{\left(-1\right)^2-4 \cdot {{1}\over{4}} \cdot -5}}{2 \cdot {{1}\over{4}}} \\&&\phantom{xxx}\color{blue}{\text{quadratic formula entered}}\\
x=2-2\cdot \sqrt{6} &\vee& x=2\cdot \sqrt{6}+2 \\&&\phantom{xxx}\color{blue}{\text{calculated}}\\
\end{array}\]
The coordinates of the intersections with the #x#-axis are: #\rv{2-2\cdot \sqrt{6},0}# and #\rv{2\cdot \sqrt{6}+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{-2.9,0}# and #\rv{6.9,0}#.

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