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 =5#, #b=-3# 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:
\[\begin{array}{rclrl}
x&=& -\dfrac{-3}{2 \cdot 5} &&\phantom{xxx}\blue{\text{formula entered}}\\
&=& {{3}\over{10}} &&\phantom{xxx}\blue{\text{simplified}}\\
\end{array}\]
The #y#-value of the vertex is calculated by entering #x={{3}\over{10}}# in the formula. Which gives:
\[\begin{array}{rclrl}
y&=& 5 \cdot {{3}\over{10}}^2 -3 \cdot {{3}\over{10}} -5
&&\phantom{xxx}\blue{\text{formula entered}}\\
&=& -{{109}\over{20}} &&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the vertex are: #\rv{{{3}\over{10}},-{{109}\over{20}}}#. To draw the point in the graph, we have to write the coordinates as decimal numbers (rounded to 1 decimal). That gives: #\rv{0.3,-5.4}#.

The intersections with the #x#-axis are the points that correspond to #y=0#.
\[\begin{array}{rcl}
5\cdot x^2-3\cdot x-5 &=& 0 \\&&\phantom{xxx}\blue{\text{the equation that should be calculated}}\\
x=\dfrac{-{-3}-\sqrt{\left(-3\right)^2-4 \cdot 5 \cdot -5}}{2 \cdot 5} &\vee& x=\dfrac{-{-3}+\sqrt{\left(-3\right)^2-4 \cdot 5 \cdot -5}}{2 \cdot 5} \\&&\phantom{xxx}\blue{\text{quadratic formula entered}}\\
x={{3-\sqrt{109}}\over{10}} &\vee& x={{\sqrt{109}+3}\over{10}} \\&&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the intersections with the #x#-axis are: #\rv{{{3-\sqrt{109}}\over{10}},0}# and #\rv{{{\sqrt{109}+3}\over{10}},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{1.3,0}#.

The four points in the graph are: #\rv{0,-5}#, #\rv{{{3}\over{10}},-{{109}\over{20}}}#, #\rv{{{3-\sqrt{109}}\over{10}},0}# and #\rv{{{\sqrt{109}+3}\over{10}},0}#.