$wk

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 =$j#, #b=$l# and #c=$n#. Seeing as the graph is a .

The intersection with the #y#-axis is equal to the value of the constant in the quadratic formula, which is equal to #$r#. That means that the coordinates of the intersection point with the #y#-axis are #\rv{0,$r}#.

The #x#-value of the vertex is given by #x=-\dfrac{b}{2 \cdot a}# and is equal to:
\[\begin{array}{rclrl}
x&=& -\dfrac{$l}{2 \cdot $j} &&\phantom{xxx}\blue{\text{formula entered}}\\
&=& $t &&\phantom{xxx}\blue{\text{simplified}}\\
\end{array}\]
The #y#-value of the vertex is calculated by entering #x=$t# in the formula. Which gives:
\[\begin{array}{rclrl}
y&=&
&&\phantom{xxx}\blue{\text{formula entered}}\\
&=& $v &&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the vertex are: #\rv{$t,$v}#. $xc

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

The four points in the graph are: #\rv{0,$r}#, #\rv{$t,$v}#, #\rv{$wd,0}# and #\rv{$wf,0}#.