Drawing of parabolas

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.

SymmetryWhen doing step 4, you can limit the number of points we need to calculate by using symmetry. A parabola is symmetrical, it is reflected in a vertical line through the vertex.

See the graph that belongs to the following formula:
$y=4\cdot x^2+4\cdot x-4$
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 $a =4$ , $b=4$ 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:
$\begin{array}{rclrl} x&=& -\dfrac{4}{2 \cdot 4} &&\phantom{xxx}\blue{\text{formula entered}}\\ &=& -{{1}\over{2}} &&\phantom{xxx}\blue{\text{simplified}}\\ \end{array}$
The $y$ -value of the vertex is calculated by entering $x=-{{1}\over{2}}$ in the formula. Which gives:
$\begin{array}{rclrl} y&=& 4 \cdot \left(-{{1}\over{2}}\right)^2 +4 \cdot -{{1}\over{2}} -4 &&\phantom{xxx}\blue{\text{formula entered}}\\ &=& -5 &&\phantom{xxx}\blue{\text{calculated}}\\ \end{array}$
The coordinates of the vertex are: $\rv{-{{1}\over{2}},-5}$ . To draw the point in the graph, we have to write the coordinates as decimal numbers (rounded to 1 decimal). That gives: $\rv{-0.5,-5}$ .

The intersections with the $x$ -axis are the points that correspond to $y=0$ .
$\begin{array}{rcl} 4\cdot x^2+4\cdot x-4 &=& 0 \\&&\phantom{xxx}\blue{\text{the equation that should be calculated}}\\ x=\dfrac{-{4}-\sqrt{4^2-4 \cdot 4 \cdot -4}}{2 \cdot 4} &\vee& x=\dfrac{-{4}+\sqrt{4^2-4 \cdot 4 \cdot -4}}{2 \cdot 4} \\&&\phantom{xxx}\blue{\text{quadratic formula entered}}\\ x={{-\sqrt{5}-1}\over{2}} &\vee& x={{\sqrt{5}-1}\over{2}} \\&&\phantom{xxx}\blue{\text{calculated}}\\ \end{array}$
The coordinates of the intersections with the $x$ -axis are: $\rv{{{-\sqrt{5}-1}\over{2}},0}$ and $\rv{{{\sqrt{5}-1}\over{2}},0}$ . To draw the point in the graph, we have to write the coordinates as decimal numbers (rounded to 1 decimal). That gives: $\rv{-1.6,0}$ and $\rv{0.6,0}$ .

The four points in the graph are: $\rv{0,-4}$ , $\rv{-{{1}\over{2}},-5}$ , $\rv{{{-\sqrt{5}-1}\over{2}},0}$ and $\rv{{{\sqrt{5}-1}\over{2}},0}$ .

The formula is already written in the form of $a \cdot x^2+b \cdot x +c$ with $a =4$ , $b=4$ , and $c=-4$ . 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.

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