Package of problems

Solutions

Can you find an equation for this parabola?

• In how many ways can you choose to show this equation?
• Which format did you initially choose and why?

Given that the diagram shows a parabola, I know that I am looking for a quadratic equation.

There are three forms in which I could choose to write a quadratic equation:

$$$y=ax^2 + bx + c\label{eq:1}$$$ $$$y=a(x-d)(x-e)\label{eq:2}$$$ $$$y=a(x-f)^2+g\label{eq:3}$$$

In order to write the equation in form $\eqref{eq:1}$ I need to know three pieces of information (because there are three unknowns). Any three different coordinate points on the parabola would work as this would enable me to write three equations in $a, b, c$ which I could then solve simultaneously.

To write the equation directly in form $\eqref{eq:2}$ (fully factorised form) I also need to know three pieces of information. This time however I specifically need to know the location of the $x$-intercepts (assuming that there are any), which will give me the values of $d$ and $e$. Given another point on the parabola I can then identify the value of $a$.

Form $\eqref{eq:3}$ (completed square form) will require a different three pieces of information. The location of the minimum or maximum point (the vertex) of the parabola, $(f, g)$, plus one additional point on the parabola to identify the value of $a$.

I can now consider the specific case of the red parabola presented above. Attempting to identify the equation of this parabola in each of the three forms described will allow me to compare efficiency.

For this particular parabola form $\eqref{eq:2}$, factorised form, is probably the most efficient form to aim for. The $x$-intercepts are clear from the graph and there are several other integer coordinates visible that can be used to identify the scaling factor, $a$.

What is the same and what is different about your approach to finding the equation of the two parabolas below, compared to the example above?

First I will consider the blue parabola.

For this parabola form $\eqref{eq:3}$, completed square form, is the more efficient form. The coordinates of the minimum point can be read directly from the graph and an integer $y$-intercept makes for a more straight-forward calculation to confirm the scaling factor, $a$. It is worth noting that the method of translating the parabola discussed in form $\eqref{eq:2}$ is also an efficient approach here.

Now I will consider the purple parabola.

For this parabola form $\eqref{eq:2}$ and form $\eqref{eq:3}$ are equivalent and more efficient that the approach used in form $\eqref{eq:1}$.

• What properties do the three parabolas have in common?

The three parabolas have the following properties in common:

• A vertical line of symmetry passing through the maximum or minimum point.

• Real roots - two of the parabolas cross the $x$-axis in two places, one touches the $x$-axis.

• A $y$-intercept that can be read from the diagram.

• At least three points that pass through integer coordinates that can be read from the graph.

• Is there an approach to finding the equation that works for all three of these parabolas, and indeed for any parabola you could be given?

• Which approach to finding the equation was the most efficient? Is it the same approach for each example?

In general all of the approaches discussed here rely on the ability to read information directly from the graph.

It is possible to use three simultaneous equations to find the equation of each parabola in form $\eqref{eq:1}$. However, this is a laborious approach and consequently there are plenty of opportunities for “silly” arithmetical errors! This is certainly not the most efficient approach!

Form $\eqref{eq:2}$, factorised form, is a very efficient approach for the red and purple parabolas. Although it may seem that this approach cannot work for the blue parabola (and indeed any parabola which doesn’t cross the $x$-axis), it may be possible to consider a translation of the parabola and to obtain the equation from there, as demonstrated in the approach to form $\eqref{eq:2}$ (for the blue parabola). This is still more efficient than attempting the original approach for form $\eqref{eq:1}$.

Form $\eqref{eq:3}$, completed square form, is a very efficient approach for all three parabolas as the coordinates of the minimum or maximum point (the vertex) of each parabola can be read directly from the graph.