Graph a Parabola

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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