What’s the same and what’s different about the following pairs of equations?

\(y=x^2-4\) and \(y+x^2-4=0\)

Let’s consider the first pair. Rearranging the second equation gives \(y=-x^2+4\). The difference between this and the first equation is that the signs are reversed on the right-hand side.

Both equations are quadratic and therefore represent parabolas when plotted on a coordinate grid. Thinking about the properties of the parabolas we can immediately see that the \(y\)-intercepts are at \(y=4\) and \(y=-4\) and that the first has a positive coefficient of \(x^2\) and the second has a negative coefficient of \(x^2\) so will be an ‘upside down’ parabola.

If we factorise both equations then we can find their roots. Factorising the first equation gives \(y=x^2-4=(x+2)(x-2)\) so it has roots at \(x= \pm 2\). Similarly, the second equation \(y=-x^2+4=-(x+2)(x-2)\). It also has its roots at \(x=2\) and \(x=-2\).

We have factorised the

*difference of two squares*.So the combined image of the two parabolas is symmetrical about both the \(x\) and \(y\)-axes.

To summarise, the two equations both represent parabolas that have the same roots. The equations represent two different parabolas as the coefficient of \(x^2\) and the \(y\)-intercept both differ.

\(y=4-9x^2\) and \(y+4=9x^2\)

If we now consider the second pair of equations we find a similar result. Rearranging the second equation gives \(y=9x^2-4\) which is the first equation with the signs reversed on the right-hand side.

Both equations are quadratic and can be represented as parabolas with a \(y\)-intercept at \(y=4\) and \(y=-4\) respectively. Factorising both equations gives \(y=4-9x^2=(2+3x)(2-3x)\) and \(y=9x^2-4=(3x+2)(3x-2)\). We can see that the (common) roots of both equations are at \(x=-\frac{2}{3}\) and \(x=\frac{2}{3}\).

The combined image of the two parabolas is once again symmetrical about both the \(x\) and \(y\)-axes.

To summarise, the two equations both represent parabolas that have the same roots. The equations represent two different parabolas as the coefficient of \(x^2\) and the \(y\)-intercept both differ.

\(2y=4x^2-6\) and \(-(y-3)=2x^2\)

Finally we shall consider the third pair of equations. Rearranging the first gives \(y=2x^2-3\) and the second \(y=-2x^2+3\).

Both equations are quadratic and can be represented as parabolas with a \(y\)-intercept at \(y=3\) and \(y=-3\) respectively.

Factorising both equations gives \[y=2x^2-3=(\sqrt{2}x+\sqrt{3})(\sqrt{2}x-\sqrt{3})\] and \[y=-2x^2+3=(-\sqrt{2}x+\sqrt{3})(\sqrt{2}x+\sqrt{3}).\]

We can see that the (common) roots of both equations are at \(x=-\frac{\sqrt{3}}{\sqrt{2}}\) and \(x=\frac{\sqrt{3}}{\sqrt{2}}\).

The combined image of the two parabolas is once again symmetrical about both the \(x\) and \(y\)-axes. To summarise, the two equations both represent parabolas that have the same roots. The equations represent two different parabolas as the coefficient of \(x^2\) and the \(y\)-intercept both differ.

All three pairs of equations share the property that they are quadratic and, when represented graphically, each pair of equations produces a pair of parabolas. The combined image of each pair is symmetrical about the \(x\) and \(y\)-axes. Each equation is distinct because of differences in the coefficients of \(x^2\) and the values of the \(y\)-intercept.

If you were given the following picture, could you suggest the equation of its pair?

From the diagram we can see that the parabola has roots at \(\pm 2\sqrt{3}\). From the shape of the parabola we can see that the coefficient of \(x^2\) will be positive.

So far this gives us that \(y=a(x+\sqrt{12})(x-\sqrt{12})\) where \(a>0\). Expanding the brackets and simplifying this expression gives \(y=ax^2-12a\). From the diagram we know that the \(y\)-intercept is at \(y=-12\). Substituting into our equation shows that when \(x=0\), \(y=-12a\), so we deduce that \(a=1\).

We have expanded the *difference of two squares*.

Now that we know the equation of the parabola in the diagram we can find its pair. We know that it will have the same roots but a \(y\)-intercept at \(y=12\) to create symmetry in the \(x\)-axis. This means that \(a=-1\) in the equation above, giving \(y=-x^2+12\).

How about if you were given this picture instead? Could you give the equation of its pair?

This picture shows another parabola but instead with the \(x\)-axis is its line of symmetry.

This suggests that the equation is of the form \(x=ay^2+by+c\). It would seem logical to treat this problem in exactly the same way as that above. We can see roots at \(y=\pm \frac{1}{3}\) and this suggests an equation of the form \(x=a\left(y+\frac{1}{3}\right)\left(y-\frac{1}{3}\right)\).

From the graph we can see that the \(x\)-intercept is at \(x=-1\). In order to achieve this we need \(a=9\) so that the equation is \(x=(3y+1)(3y-1)=9y^2-1\).

Now to find its pair. We know that it will have the same roots but an \(x\)-intercept at \(x=1\). Therefore the paired parabola will be given by the equation \(x=-(3y+1)(3y-1)=1-9y^2\).