Quadratic equations: Drawing parabolas
Drawing of parabolas
We have seen that the graph of a quadratic is a parabola. We have also seen how the intersection points with the axes, the vertex and other points with particular values of #x# of the parabola can be calculated. From these calculated values we can easily draw the graph of a quadratic.
Procedure drawing parabola
Procedure |
geogebra plaatje
|
|
We will draw the graph of a quadratic. |
||
Step 1 |
Determine the intersection point with the #y#-axis. |
|
Step 2 |
Determine the vertex. |
|
Step 3 |
Determine the intersection points with the #x#-axis, if there are any. |
|
Step 4 |
Substitute values for #x# in the formula in such a way that we have at least 4 points we can draw. |
|
Step 5 |
Draw these points in the coordinate system and connect them by a smooth parabola. |
\[y=3\cdot x^2+2\cdot x-3\]
Draw the intersection with the #y#-axis, the vertex, and the intersections with the #x#-axis.

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 =3#, #b=2# and #c=-3#. 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 #-3#. That means that the coordinates of the intersection point with the #y#-axis are #\rv{0,-3}#.
The #x#-value of the vertex is given by #x=-\dfrac{b}{2 \cdot a}# and is equal to:
\[\begin{array}{rclrl}
x&=& -\dfrac{2}{2 \cdot 3} &&\phantom{xxx}\blue{\text{formula entered}}\\
&=& -{{1}\over{3}} &&\phantom{xxx}\blue{\text{simplified}}\\
\end{array}\]
The #y#-value of the vertex is calculated by entering #x=-{{1}\over{3}}# in the formula. Which gives:
\[\begin{array}{rclrl}
y&=& 3 \cdot \left(-{{1}\over{3}}\right)^2 +2 \cdot -{{1}\over{3}} -3
&&\phantom{xxx}\blue{\text{formula entered}}\\
&=& -{{10}\over{3}} &&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the vertex are: #\rv{-{{1}\over{3}},-{{10}\over{3}}}#. 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,-3.3}#.
The intersections with the #x#-axis are the points that correspond to #y=0#.
\[\begin{array}{rcl}
3\cdot x^2+2\cdot x-3 &=& 0 \\&&\phantom{xxx}\blue{\text{the equation that should be calculated}}\\
x=\dfrac{-{2}-\sqrt{2^2-4 \cdot 3 \cdot -3}}{2 \cdot 3} &\vee& x=\dfrac{-{2}+\sqrt{2^2-4 \cdot 3 \cdot -3}}{2 \cdot 3} \\&&\phantom{xxx}\blue{\text{quadratic formula entered}}\\
x={{-\sqrt{10}-1}\over{3}} &\vee& x={{\sqrt{10}-1}\over{3}} \\&&\phantom{xxx}\blue{\text{calculated}}\\
\end{array}\]
The coordinates of the intersections with the #x#-axis are: #\rv{{{-\sqrt{10}-1}\over{3}},0}# and #\rv{{{\sqrt{10}-1}\over{3}},0}#. To draw the point in the graph, we have to write the coordinates as decimal numbers (rounded to 1 decimal). That gives:#\rv{-1.4,0}# and #\rv{0.7,0}#.
The four points in the graph are: #\rv{0,-3}#, #\rv{-{{1}\over{3}},-{{10}\over{3}}}#, #\rv{{{-\sqrt{10}-1}\over{3}},0}# and #\rv{{{\sqrt{10}-1}\over{3}},0}#.
The requested points are connected by a smooth curve in the figure: the parabola that opens upward is given by the formula.
Or visit omptest.org if jou are taking an OMPT exam.