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



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


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


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).


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



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:


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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