Can you find the correct one of graphs M to X to go in each cell of the grid below?

In order to match up the graphs with their equations, we need to identify some of the features of each curve, such as

- the location of vertical asymptotes,
- where it crosses the \(x\) and \(y\)-axes,
- whether the function is positive or negative for ranges of \(x\) values,
- its behaviour as \(x\) becomes very large in positive and negative directions.

To find some of these features, it can be helpful to rearrange the algebra or to consider the effect of simple graph transformations. Here are a few examples of things you might have thought of.

### Finding \(\frac{a(x)}{b(x)}\) for \(b(x)=2x-1\)

We want to identify the graph of \(y=\dfrac{x+1}{2x-1}\).

We can see that it will have an \(x\)-intercept at \((-1,0)\) and a \(y\)-intercept at \((0,-1)\). It will also have a vertical asymptote at \(x=\frac{1}{2}\). As \(x\) increases from just less than \(\frac{1}{2}\) to just more than \(\frac{1}{2}\), the denominator goes from negative to positive so our graph will do the same at the asymptote.

When \(x\) is very large, either positive or negative, we have \(y\approx\dfrac{x}{2x}=\dfrac{1}{2}\).

The shape of the curve is not immediately obvious, but we can do some rearranging of the algebra. \[y=\frac{x+1}{2x-1}=\frac{\frac{1}{2}(2x-1)+\frac{3}{2}}{2x-1} = \frac{1}{2}+\frac{3}{2}\frac{1}{2x-1} = \frac{1}{2} +\frac{3}{4}\frac{1}{(x-1/2)}\] So now we can see that our function is a hyperbola, \(\frac{1}{x}\), that has been translated horizontally by \(\frac{1}{2}\), scaled vertically by \(\frac{3}{4}\) and then translated up by \(\frac{1}{2}\).

This confirms that as \(x\to\pm\infty\), \(y\to\frac{1}{2}\) and the correct graph is (M).

### Finding \(\frac{1}{b(x)}\) for \(b(x)=x^2-5x+8\)

It is helpful to first explore the features of the quadratic, \(b(x)\). It doesn’t factorise easily, so let’s try completing the square. \[x^2-5x+8 = \left(x-\tfrac{5}{2}\right)^2 + 8-\tfrac{25}{4} = \left(x-\tfrac{5}{2}\right)^2 + \tfrac{7}{4}\] So we have that

- \(b(x)\) is a parabola with a minimum at \(\left(\frac{5}{2},\frac{7}{4}\right)\), so \(y=\frac{1}{b(x)}\) has a maximum at \(\left(\frac{5}{2},\frac{4}{7}\right)\);
- \(b(x)\) is always positive, so \(y=\frac{1}{b(x)}\) is also always positive;
- \(b(x)\) has no zeros, so \(y=\frac{1}{b(x)}\) has no vertical asymptotes;
- as \(x\to\pm\infty\), \(\frac{1}{b(x)}\to 0\).

The correct graph is (S).

### Finding \(\frac{b(x)}{a(x)}\) for \(b(x)=x^2-5x+8\)

We want to find \(y=\dfrac{x^2-5x+8}{x+1}\). We discovered above that the quadratic numerator has no zeros, so this curve does not cross the \(x\)-axis.

It has a \(y\)-intercept at \(8\), and a vertical asymptote at \(x=-1\).

Let’s see what happens if we do an algebraic division.

Using long division, \[ \begin{array}{rll} \phantom{x^2+5}x-6 & \text{ r }14 \cmlongdivbr x+1 \cmlongdiv{x^2-5x+8} \cmlongdivbr \underline{x^2+\phantom{0}x}\phantom{{}+8} && \hbox{[$x \times (x+1)$]} \cmlongdivbr \phantom{x^2}{}-6x+8 && \hbox{[subtract...]} \cmlongdivbr \phantom{x^2}\underline{{}-6x-6} && \hbox{[$-6(x+1)$]} \cmlongdivbr \phantom{x^2-5x+{}}14 && \hbox{[remainder]} \cmlongdivbr \end{array} \]

or using a grid method, \[ \begin{array}{r|rr} & \phantom{0}x\phantom{x^2}-6 \\ \hline x & x^2-6x & 14 \\ +1 & x\phantom{x^2}-6 \end{array} \]

The remainder of \(14\) tells us that \(x^2-5x+8=(x+1)(x-6)+14\).

So we have \(y=x-6 +\frac{14}{x+1}\) and the curve is a hyperbola added to a straight line of gradient \(1\). This gives the perhaps surprising result that as \(x\to\pm\infty\), \(y\to x-6\).

The correct graph is (U).

### Finding the third row graphs

The quadratic in the third row, \(b(x)=x^2-5x-6\), looks very similar to that in the second row. However, this one factorises such that \(b(x)\) has zeros at \(x=-1\) and \(x=6\). This gives rise to some vertical asymptotes, but also allows some cancelling to occur in the algebraic fractions.