### Two points on the axis

Using GeoGebra, plot two points \(A\) and \(B\) on the \(x\)-axis of a graph. Can you instruct GeoGebra to draw a quadratic which passes through both of them?

Here is a solution:

We want the quadratic to pass through \(A\) and \(B\) on the \(x\)-axis. If we write the coordinates of these points as \(A=(a,0)\) and \(B=(b,0)\), we see that the roots of the quadratic must be \(a\) and \(b\).

A quadratic which has these roots is \(y=(x-a)(x-b)\), so this is what we need.

To draw this in GeoGebra, we need to work out \(a\) and \(b\). We can do this by typing:

```
a = x(A)
b = x(B)
y = (x-a)(x-b)
```

or all in one go as:

`y = (x-x(A))(x-x(B))`

#### Taking it further

Developing this idea further, add a point \(C\) to the \(y\)-axis. Can you make your quadratic pass through \(A\), \(B\) and \(C\), like this?

Here is a solution:

In the first part of this problem, we found that one quadratic which passes through \(A\) and \(B\) has equation \(y=(x-a)(x-b)\).

We could expand this to get \(y=x^2-(a+b)x+ab\), and we see that the coefficient of \(x^2\) is \(1\).

There was no need for us to make this choice, though: we could choose any (non-zero) coefficient for \(x^2\). So we could equally well have used \(y=k(x-a)(x-b)\) as our equation, as this also passes through \(A\) and \(B\) on the \(x\)-axis.

This new quadratic crosses the \(y\)-axis where \(x=0\); substituting in \(x=0\) shows that the intercept is at \((0,kab)\). As we want it to cross the \(y\)-axis at \(C(0,c)\), we choose \(k\) so that \[kab=c.\] This can easily be rearranged to get \(k=\dfrac{c}{ab}\).

We can therefore specify the whole quadratic to GeoGebra as

`y = y(C)/(x(A)*x(B)) * (x-x(A))(x-x(B))`

We are assuming here that \(ab\ne0\). If \(ab=0\), this means that either \(A\) or \(B\) lies at the origin, and it is not going to be possible for the quadratic to then pass through both \(A\) and \(B\) along with \(C\).

The one exception is if \(C\) also lies at the origin. In this case, we have \(k=0/0\), which is not defined. And geometrically, in this situation, we are only specifying where the quadratic crosses the \(x\)-axis, so there are many possibilities for the quadratic.

### One point as the vertex

On a new graph, plot a point \(A\). Can you instruct GeoGebra to draw a quadratic which has its vertex at \(A\)?

Here is a solution:

Since we are thinking about the vertex of the quadratic, it makes most sense to work with the completed square form of the quadratic: \[y=(x-a)^2+c.\] This has its vertex at \((a,c)\).

So if we write the coordinates of point \(A\) as \((a_1, a_2)\), our quadratic could be \[y=(x-a_1)^2+a_2.\]

In GeoGebra terms, we can type this in as either:

```
a1 = x(A)
a2 = y(A)
y=(x-a1)^2+a2
```

or all in one go as

`y=(x-x(A))^2+y(A)`

#### Taking it further

Developing this idea further, add a point \(B\) to the \(y\)-axis. Can you make your quadratic have its vertex at \(A\) and pass through \(B\), like this?

Here is a solution:

In the first part of this problem, we found that one quadratic which passes through \(A\) has equation \(y=(x-a_1)^2+a_2\).

As we noted in the “taking it further” part of question 1 above, there was no need for us to make the coefficient of \(x^2\) equal to \(1\). We could equally well have used \[y=k(x-a_1)^2+a_2\] for any choice of \(k\), for this also has its vertex at \(A(a_1,a_2)\).

This new quadratic crosses the \(y\)-axis where \(x=0\), so at \((0,ka_1^2+a_2)\). As we want it to cross the \(y\)-axis at \(B(0,b_2)\) (where \(B\) has coordinates \((0,b_2)\), we will choose \(k\) so that \[ka_1^2+a_2=b_2.\] This can easily be rearranged to get \(k=\dfrac{b_2-a_2}{a_1^2}\).

We can specify this to GeoGebra all in one go as

`y = (y(B)-y(A))/(x(A)^2) * (x-x(A))^2 + y(A)`

As in question 1, we are assuming that \(a_1\ne0\). What would happen if \(a_1=0\)?

### Any three points

The final (very hard) challenge: plot three points \(A\), \(B\) and \(C\) anywhere on the graph, and draw a quadratic passing through these three points, like this:

Here is a solution:

We write \(A(a_1,a_2)\), \(B(b_1,b_2)\) and \(C(c_1,c_2)\) as in the suggestion.

Is there more than one quadratic you could draw?