Point By Point

We can draw the graph of a parabola by taking all the integer values of $x$ shown on the $x$ axis, calculate the corresponding $y$ value for each, plot each point in turn and join them with a smooth curve.

This method is not very efficient.

Using Special Points

Another way is to calculate the special points of the parabola and plot only those.

The special points are:

$y$ -intercept
vertex
$x$ -intercepts (if there are any)
symmetry of the $y$ -intercept

The special points can be calculated quite easily from the quadratic expression given in factored form or in completed square form.

Special Points from Factored Form

Example

$y=(x-3)(x+5)$

The $y$ intercept occurs when the $x$ coordinate is zero. Therefore when $x=0$ :

$y=(x-3)(x+5)=(0-3)(0+5)=(-3)(5)=-15$

The $y$ -intercept is the point $(0,-15)$ .

The $x$ -intercepts occur when the $y$ coordinate is zero. To locate them, we need to solve the equation

$0=(x-3)(x+5)$

When any two numbers multiply to give zero, one or the other or both of the numbers must be zero. Therefore, if

$\begin{align*}0&=(x-3)(x+5)\\[10 pt] 0&=x-3 \text{ or } & 0&=x+5\\[10 pt] \Rightarrow x&=3 &\Rightarrow x&=-5\end{align}$

The $x$ -intercepts occur at the points $(3,0)$ and $(-5,0)$ .

Vertex: The $x$ coordinate of the vertex is right in the middle of the intercepts. To find the middle of $-5$ and $3$ , we can do a variety of calculations. One way is to add them together and divide by 2 (find the average of the two numbers).

$x_{\text{vertex}}=\dfrac{-5+3}{2}=\dfrac{-2}{2}=-1$

Now to find the $y$ coordinate of the vertex, we simply substitute the $x$ coordinate into the equation for $y$ . When $x=-1$ :

$y=(x-3)(x+5)=(-1-3)(-1+5)=(-4)(4)=-16$

$y_{\text{vertex}}=-16$

The coordinates of the vertex are $(-1,-16)$ .

Let’s plot the points $(0,-15),\,(-5,0),\,(3,0),\,(-1,-16)$ :

The last point to plot is the symmetry of the $y$ intercept. If you imagine a mirror line through the vertex, and reflect the $y$ intercept over that to plot a fifth point.

We are now ready to draw a smooth curve through the plotted points, making sure to round off the vertex and not make it too pointy.

Special Points from Completed Square form:

We calculate the same special points, however our calculations look slightly different as our equation is presented differently.

Let’s graph the parabola $y=(x-3)^2-8$ .

The $y$ -intercept: The point on the parabola that lies on the $y$ -axis has $x=0$ . We substitute $x=0$ into the equation:

$\begin{align*}y&=(x-3)^2-8\\[10 pt]&=(0-3)^2-8\\[10 pt]&=(-3)^2-8\\[10 pt]&=9-8=1\end{align}$

The $y$ -intercept has coordinates $(0,1)$

The $x$ -intercepts: We substitute $y=0$ and solve directly, remembering that when we take a square root we must remember to consider both the positive and the negative square root. For example, if we know that $x^2=25$ we know that both $-5$ and $+5$ are solutions.

$\begin{align*}y&=(x-3)^2-8\\[10 pt]0&=(x-3)^2-8\\[10 pt]8&=(x-3)^2\end{align}$

Now we take the square root and consider both negative and positive:

$\begin{align*}-\sqrt{8}&=x-3 &+\sqrt{8}&=x-3\\[10 pt]-\sqrt{8}+3&=x &\sqrt{8}+3&=x\\[10 pt]x&=0.17&x&=5.83\quad\text{2 d.p.}\end{align}$

Note that the exact value of these roots can be simplified:

$\pm\sqrt{8}+3=\pm\sqrt{4\times2}+3=\pm2\sqrt{2}+3$

The $x$ -intercepts have coordinates $(0.17,0)$ and $(5.83,0)$ .

The vertex: The lowest point on the parabola has the lowest $y$ value. The lowest square number is zero. Since $y=(\text{something})^2-8$ , we require ‘something’ to be equal to zero.

In our case, $y=(x-3)^2-8$ . Solve $x-3=0$ . The vertex has the $x$ coordinate $x=3$ .

To calculate the $y$ coordinate, as for any $y$ coordinate, we substitute the $x$ coordinate into the equation:

$\begin{align*}y&=(x-3)^2-8\\[10 pt]y&=(3-3)^2-8\\[10 pt]y&=0-8=-8\end{align*}$

The coordinates of the vertex are $(3,-8)$ .

Let’s plot the points $(0,1)$ , $(0.17,0)$ , $(5.83,0)$ and $(3,-8)$ on the axes:

…and draw the parabola:

Graph from standard form $y=ax^2+bx+c$

To graph directly from the standard form, we need to employ processes that occur later in this unit. In summary:

To calculate the $y$ intercept, again we set $x=0$ and find that $y=a(0)^2+b(0)+c=c$ . Therefore the $y$ intercept has coordinates $(0,c)$ .

To calculate the $x$ intercepts, we need to use one of the following processes:

To calculate the $x$ coordinate of the vertex, we need to use one of these processes:

use $x_{\text{vertex}}=\frac{x_1+x_2}{2}$ once the $x$ intercepts are known;
use the formula $x_{\text{vertex}}=-\frac{b}{2a}$ ;
write the expression in completed square form

We then substitute the $x$ coordinate into the equation to find the $y$ coordinate of the vertex.

These processes are discussed in the rest of the unit.

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