Mapping a function

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Other multiples

Here are mapping diagrams of our original \(f(x)=3x\) together with the other functions given. We have decided to centre each of the input numberlines on zero and use a scale of one unit per tick mark; as we discovered in the Warm-up, this makes no difference to the mapping diagram, only to the labelling.

Clearly, the multiplier \(a\) in \(f(x)=ax+b\) makes a significant difference to the mapping diagrams:

• \(f(x)=2x\) is less “spread out” than \(f(x)=3x\), and \(f(x)=\frac{1}{2}x\) results in the arrows becoming closer together on the output numberline.
• \(f(x)=x\) consists of parallel arrows.
• \(f(x)=-x\) and \(f(x)=-2x\) both lead to arrows which cross themselves. (Incidentally, this is one way to see that multiplying an inequality by a negative number reverses the inequality.)

As we saw in the Warm-up, though, the mapping diagram for \(f(x)=ax+b\) remains the same if we change the “\(+b\)” part (as long as we stick to our conventions).

Extending the arrows

Here are the mapping diagrams for the same six functions as above, with extended lines.

The extended lines all meet at a single point (except for the function \(f(x)=x\)), which we could think of as a centre of enlargement or a focal point. This means that we can describe a linear function as a scaling of the left-hand numberline into the right-hand one.

Some further questions:

• Can you prove that every linear function behaves in this way: that it is a scaling of the left-hand numberline into the right-hand one?

• Can you find a rule for the location of the focal point for \(f(x)=ax+b\)?

Solution

If we can find a focal point for all of the lines, then we will have answered most of the first question as well.

The easiest case for calculations is when we have a function \(f(x)=ax\) (so \(b=0\)), and we centre our numberlines on the arrow from \(0\) to \(0\). (And we know from earlier that this doesn’t change what the mapping diagram looks like.) If we put a coordinate system on our diagram, and imagine that our numberlines cross the \(x\)-axis at the origin and \((1,0)\), then our diagram looks like this:

An arrow then goes from the point with coordinates \((0,k)\), which represents some value of function input, to \((1,ak)\). It doesn’t even matter what the scale is on the numberlines: the arrow will still go from \((0,k)\) to \((1,ak)\) for some value of \(k\).

The gradient of this line is then \(\dfrac{ak-k}{1-0}=ak-k\), and so it has equation (in our “normal” \(x\)-\(y\) coordinate system) \(y=(ak-k)x+k\). This line will meet the \(x\)-axis when \(y=0\), so \(x=\dfrac{-k}{ak-k}=\dfrac{1}{1-a}\). This does not depend on \(k\), so every line must pass through the point \(\left(\dfrac{1}{1-a},0\right)\).

This means that the lines all pass through a single focal point.

But we have already seen that this is not true for the function \(f(x)=x\). Where does the argument fail in this case?

Using this formula, we can also say something about the location of the focal point, depending on the value of \(a\):

Value of \(a\)	Location of focal point
\(a>1\)	to left of left-hand (input) numberline
\(0<a<1\)	to right of right-hand (output) numberline
\(a<0\)	between the numberlines
\(a=0\)	on the right-hand (output) numberline

This also shows that the transformation from the left-hand (input) numberline to the right-hand (output) numberline is a scaling: the scale factor is clearly just \(a\) from the same diagram, as the line segment from \((0,0)\) to \((0,k)\) becomes the line segment from \((1,0)\) to \((1,ak)\).

The same argument also works if we consider the more general case where we centre the diagram on a different arrow, or if we have a function \(f(x)=ax+b\) instead. Can you see why?

Taking it further: changing the conventions

Here are the mapping diagrams for \(f(x)=3x+1\) with the left one centred as we have been doing, and the right one with both numberlines centred at \(0\). For each one, we have drawn five arrows starting at \(-1\), \(-\frac{1}{2}\), \(0\), \(\frac{1}{2}\) and \(1\).

Some observations we might make:

• The second diagram looks fairly similar to the first in many ways.

• The focal point is the same distance away from the left-hand numberline in both diagrams.

• The second diagram looks just like our earlier diagrams if we imagine the left-hand numberline being centred on \(-\frac{1}{2}\).

• The second diagram looks like the first, just with the right-hand numberline moved up by \(1\) unit and everything else pulled along with it. (The technical name for this transformation is a shear.)