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. 
See the graph that belongs to the following formula:
\[y=2\cdot x^2+3\cdot x2\]
Draw the intersection with the #y#axis, the vertex, and the intersections with the #x#axis.
\[y=2\cdot x^2+3\cdot x2\]
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 followed:
The formula is already written in the form of #a \cdot x^2+b \cdot x +c# with #a =2#, #b=3# and #c=2#. 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 #2#. That means that the coordinates of the intersection point with the #y#axis are #\rv{0,2}#.
The #x#value of the vertex is given by #x=\dfrac{b}{2 \cdot a}# and is equal to:
\[\begin{array}{rclrl}
x&=& \dfrac{3}{2 \cdot 2} &&\phantom{xxx}\color{blue}{\text{formula entered}}\\
&=& {{3}\over{4}} &&\phantom{xxx}\color{blue}{\text{simplified}}\\
\end{array}\]
The #y#value of the vertex is calculated by entering #x={{3}\over{4}}# in the formula. Which gives:
\[\begin{array}{rclrl}
y&=& 2 \cdot \left({{3}\over{4}}\right)^2 +3 \cdot {{3}\over{4}} 2
&&\phantom{xxx}\color{blue}{\text{formula entered}}\\
&=& {{25}\over{8}} &&\phantom{xxx}\color{blue}{\text{calculated}}\\
\end{array}\]
The coordinates of the vertex are: #\rv{{{3}\over{4}},{{25}\over{8}}}#. To draw the point in the graph, we have to write the coordinates as decimal numbers (rounded to 1 decimal). That gives: #\rv{0.8,3.1}#.
The intersections with the #x#axis are the points that correspond to #y=0#.
\[\begin{array}{rcl}
2\cdot x^2+3\cdot x2 &=& 0 \\&&\phantom{xxx}\color{blue}{\text{the equation that should be calculated}}\\
x=\dfrac{{3}\sqrt{3^24 \cdot 2 \cdot 2}}{2 \cdot 2} &\vee& x=\dfrac{{3}+\sqrt{3^24 \cdot 2 \cdot 2}}{2 \cdot 2} \\&&\phantom{xxx}\color{blue}{\text{quadratic formula entered}}\\
x=2 &\vee& x={{1}\over{2}} \\&&\phantom{xxx}\color{blue}{\text{calculated}}\\
\end{array}\]
The coordinates of the intersections with the #x#axis are: #\rv{2,0}# and #\rv{{{1}\over{2}},0}#. To draw the point in the graph, we have to write the coordinates as decimal numbers (rounded to 1 decimal). That gives:#\rv{2,0}# and #\rv{0.5,0}#.
The four points in the graph are: #\rv{0,2}#, #\rv{{{3}\over{4}},{{25}\over{8}}}#, #\rv{2,0}# and #\rv{{{1}\over{2}},0}#.
The formula is already written in the form of #a \cdot x^2+b \cdot x +c# with #a =2#, #b=3#, and #c=2#. As #a>0# the graph is a parabola that opens upward .
The requested points are connected by a smooth curve in the figure: the parabola that opens upward is given by the formula.
The requested points are connected by a smooth curve in the figure: the parabola that opens upward is given by the formula.
Unlock full access
Teacher access
Request a demo account. We will help you get started with our digital learning environment.
Student access
Is your university not a partner?
Get access to our courses via Pass Your Math independent of your university. See pricing and more.
Or visit omptest.org if jou are taking an OMPT exam.
Or visit omptest.org if jou are taking an OMPT exam.