write down the equation of the vertical asymptote.

find the equation of the horizontal asymptote.

Find $f^{-1}\left(x\right)$.

Using an algebraic approach, show that the graph of $f^{-1}$ is obtained by a reflection of the graph of $f$ in the $y$-axis followed by a reflection in the $x$-axis.

Find the value of $p$ and the value of $q$.

Hence, find the area enclosed by the graph of $f$ and the graph of $f^{-1}$.

$x=-4$          A1

[1 mark]

attempt to substitute into $y=\frac{a}{c}$  OR  table with large values of $x$  OR  sketch of $f$ showing asymptotic behaviour          (M1)

$y=4$          A1

[2 marks]

$y=\frac{4x+1}{x+4}$

attempt to interchange $x$ and $y$ (seen anywhere)        M1

$xy+4y=4x+1$   OR   $xy+4x=4y+1$         (A1)

$xy-4x=1-4y$   OR   $xy-4y=1-4x$         (A1)

$f^{-1}\left(x\right)=\frac{1-4x}{x-4}$  (accept $y=\frac{1-4x}{x-4}$)         A1

[4 marks]

reflection in $y$-axis given by $f\left(-x\right)$         (M1)

$f\left(-x\right)=\frac{-4x+1}{-x+4}$         (A1)

reflection of their $f\left(-x\right)$ in $x$-axis given by $-f\left(-x\right)$ accept "now $-f\left(x\right)$"        M1

$\left(-f\left(-x\right)=\right) -\frac{-4x+1}{-x+4}$

$=\frac{-4x+1}{x-4}$  OR  $\frac{4x-1}{-x+4}$         A1

$=\frac{1-4x}{x-4} \left(=f^{-1}\left(x\right)\right)$         AG

Note: If the candidate attempts to show the result using a particular coordinate on the graph of $f$ rather than a general coordinate on the graph of $f$, where appropriate, award marks as follows:
M0A0 for eg $(2,3)\to (-2,3)$
M0A0 for $(-2,3)\to (-2,-3)$

[4 marks]

attempt to solve $f\left(x\right)=f^{-1}\left(x\right)$ using graph or algebraically         (M1)

$p=-1$  AND  $q=1$         A1

Note: Award (M1)A0 if only one correct value seen.

[2 marks]

attempt to set up an integral to find area between $f$ and $f^{-1}$         (M1)

$\underset{-1}{\overset{1}{\int }}\left(\frac{4x+1}{x+4}-\frac{1-4x}{x-4}\right)dx$         (A1)

$=0.675231\dots$

$=0.675$         A1

[3 marks]

Candidates mostly found the first part of this question accessible, with many knowing how to find the equation of both asymptotes. Common errors included transposing the asymptotes, or finding where an asymptote occurred but not giving it as an equation.

Candidates knew how to start part (b)(i), with most attempting to find the inverse function by firstly interchanging $x$ and $y$. However, many struggled with the algebra required to change the subject, and were not awarded all the marks. A common error in this part was for candidates to attempt to find an expression for $f\text{'}\left(x\right)$, rather than one for $f^{-1}\left(x\right)$. Few candidates were able to answer part (b)(ii). Many appeared not to know that a reflection in the $y$-axis is given by $f(-x)$, or that a reflection in the $x$-axis is given by $-f\left(x\right)$. Many of those that did, multiplied both the numerator and denominator by $-1$ when taking the negative of their $f(-x)$ , i.e. $-\left(\frac{-4x+1}{-x+4}\right)$ was often simplified as $\frac{4x-1}{x-4}$. However, the majority of candidates either did not attempt this question part or attempted to describe a graphical approach often involving a specific point, rather than an algebraic approach.

Those that attempted part (c), and had the correct expression for $f^{-1}\left(x\right)$, were usually able to gain all the marks. However, those that had an incorrect expression, or had found $f\text{'}\left(x\right)$, often proceeded to find an area, even when there was not an area enclosed by their two curves.