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:
\[y={{x^2}\over{4}}+3\cdot x+8\]
Draw the intersection with the #y#axis, the vertex, and the intersections with the #x#axis.
\[y={{x^2}\over{4}}+3\cdot x+8\]
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 ={{1}\over{4}}#, #b=3# and #c=8#. Seeing as #a<0# the graph is a parabola that opens downward.
The intersection with the #y#axis is equal to the value of the constant in the quadratic formula, which is equal to #8#. That means that the coordinates of the intersection point with the #y#axis are #\rv{0,8}#.
The #x#value of the vertex is given by #x=\dfrac{b}{2 \cdot a}# and is equal to:
\[\begin{array}{rclrl}
x&=& \dfrac{3}{2 \cdot {{1}\over{4}}} &&\phantom{xxx}\color{blue}{\text{formula entered}}\\
&=& 6 &&\phantom{xxx}\color{blue}{\text{simplified}}\\
\end{array}\]
The #y#value of the vertex is calculated by entering #x=6# in the formula. Which gives:
\[\begin{array}{rclrl}
y&=& {{1}\over{4}} \cdot 6^2 +3 \cdot 6 +8
&&\phantom{xxx}\color{blue}{\text{formula entered}}\\
&=& 17 &&\phantom{xxx}\color{blue}{\text{calculated}}\\
\end{array}\]
The coordinates of the vertex are: #\rv{6,17}#.
The intersections with the #x#axis are the points that correspond to #y=0#.
\[\begin{array}{rcl}
{{x^2}\over{4}}+3\cdot x+8 &=& 0 \\&&\phantom{xxx}\color{blue}{\text{the equation that should be calculated}}\\
x=\dfrac{{3}\sqrt{3^24 \cdot {{1}\over{4}} \cdot 8}}{2 \cdot {{1}\over{4}}} &\vee& x=\dfrac{{3}+\sqrt{3^24 \cdot {{1}\over{4}} \cdot 8}}{2 \cdot {{1}\over{4}}} \\&&\phantom{xxx}\color{blue}{\text{quadratic formula entered}}\\
x=62\cdot \sqrt{17} &\vee& x=2\cdot \sqrt{17}+6 \\&&\phantom{xxx}\color{blue}{\text{calculated}}\\
\end{array}\]
The coordinates of the intersections with the #x#axis are: #\rv{62\cdot \sqrt{17},0}# and #\rv{2\cdot \sqrt{17}+6,0}#. To draw the point in the graph, we have to write the coordinates as decimal numbers (rounded to 1 decimal). That gives:#\rv{2.2,0}# and #\rv{14.2,0}#.
The four points in the graph are: #\rv{0,8}#, #\rv{6,17}#, #\rv{62\cdot \sqrt{17},0}# and #\rv{2\cdot \sqrt{17}+6,0}#.
The formula is already written in the form of #a \cdot x^2+b \cdot x +c# with #a ={{1}\over{4}}#, #b=3#, and #c=8#. As #a<0# the graph is a parabola that opens downward .
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.
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