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=3# and #c=-4#. 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 #-4#. That means that the coordinates of the intersection point with the #y#-axis are #\rv{0,-4}#.

The #x#-value of the vertex is given by #x=-\dfrac{b}{2 \cdot a}# and is equal to:
\[\begin{array}{rclrl}
x&=& -\dfrac{3}{2 \cdot 1} &&\phantom{xxx}\blue{\text{formula entered}}\\
&=& -{{3}\over{2}} &&\phantom{xxx}\blue{\text{simplified}}\\
\end{array}\]
The #y#-value of the vertex is calculated by entering #x=-{{3}\over{2}}# in the formula. Which gives:
\[\begin{array}{rclrl}
y&=& \left(-{{3}\over{2}}\right)^2 +3 \cdot -{{3}\over{2}} -4
&&\phantom{xxx}\blue{\text{formula entered}}\\
&=& -{{25}\over{4}} &&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the vertex are: #\rv{-{{3}\over{2}},-{{25}\over{4}}}#. To draw the point in the graph, we have to write the coordinates as decimal numbers (rounded to 1 decimal). That gives: #\rv{-1.5,-6.2}#.

The intersections with the #x#-axis are the points that correspond to #y=0#.
\[\begin{array}{rcl}
x^2+3\cdot x-4 &=& 0 \\&&\phantom{xxx}\blue{\text{the equation that should be calculated}}\\
x=\dfrac{-{3}-\sqrt{3^2-4 \cdot 1 \cdot -4}}{2 \cdot 1} &\vee& x=\dfrac{-{3}+\sqrt{3^2-4 \cdot 1 \cdot -4}}{2 \cdot 1} \\&&\phantom{xxx}\blue{\text{quadratic formula entered}}\\
x=1 &\vee& x=-4 \\&&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the intersections with the #x#-axis are: #\rv{1,0}# and #\rv{-4,0}#.

The four points in the graph are: #\rv{0,-4}#, #\rv{-{{3}\over{2}},-{{25}\over{4}}}#, #\rv{1,0}# and #\rv{-4,0}#.