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{3}}#, #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:
\[\begin{array}{rclrl}
x&=& -\dfrac{4}{2 \cdot {{1}\over{3}}} &&\phantom{xxx}\color{blue}{\text{formula entered}}\\
&=& -6 &&\phantom{xxx}\color{blue}{\text{simplified}}\\
\end{array}\]
The #y#-value of the vertex is calculated by entering #x=-6# in the formula. Which gives:
\[\begin{array}{rclrl}
y&=& {{1}\over{3}} \cdot \left(-6\right)^2 +4 \cdot -6 -7
&&\phantom{xxx}\color{blue}{\text{formula entered}}\\
&=& -19 &&\phantom{xxx}\color{blue}{\text{calculated}}\\
\end{array}\]
The coordinates of the vertex are: #\rv{-6,-19}#.

The intersections with the #x#-axis are the points that correspond to #y=0#.
\[\begin{array}{rcl}
{{x^2}\over{3}}+4\cdot x-7 &=& 0 \\&&\phantom{xxx}\color{blue}{\text{the equation that should be calculated}}\\
x=\dfrac{-{4}-\sqrt{4^2-4 \cdot {{1}\over{3}} \cdot -7}}{2 \cdot {{1}\over{3}}} &\vee& x=\dfrac{-{4}+\sqrt{4^2-4 \cdot {{1}\over{3}} \cdot -7}}{2 \cdot {{1}\over{3}}} \\&&\phantom{xxx}\color{blue}{\text{quadratic formula entered}}\\
x=-\sqrt{57}-6 &\vee& x=\sqrt{57}-6 \\&&\phantom{xxx}\color{blue}{\text{calculated}}\\
\end{array}\]
The coordinates of the intersections with the #x#-axis are: #\rv{-\sqrt{57}-6,0}# and #\rv{\sqrt{57}-6,0}#. To draw the point in the graph, we have to write the coordinates as decimal numbers (rounded to 1 decimal). That gives:#\rv{-13.5,0}# and #\rv{1.5,0}#.

The four points in the graph are: #\rv{0,-7}#, #\rv{-6,-19}#, #\rv{-\sqrt{57}-6,0}# and #\rv{\sqrt{57}-6,0}#.