Consider \(f(x) = \frac{{{x^2} - 5x + 4}}{{{x^2} + 5x + 4}}\).

(a)     Find the equations of all asymptotes of the graph of f.

(b)     Find the coordinates of the points where the graph of f meets the x and y axes.

(c)     Find the coordinates of

(i)     the maximum point and justify your answer;

(ii)     the minimum point and justify your answer.

(d)     Sketch the graph of f, clearly showing all the features found above.

(e)     Hence, write down the number of points of inflexion of the graph of f.

(a)     \({x^2} + 5x + 4 = 0 \Rightarrow x = - 1{\text{ or }}x = - 4\)     (M1)

so vertical asymptotes are x = – 1 and x = – 4     A1

as \(x \to \infty \) then \(y \to 1\) so horizontal asymptote is y = 1     (M1)A1

[4 marks]

 

(b)     \({x^2} - 5x + 4 = 0 \Rightarrow x = 1{\text{ or }}x = 4\)     A1

\(x = 0 \Rightarrow y = 1\)     A1

so intercepts are (1, 0), (4, 0) and (0,1)

[2 marks]

 

(c)     (i)     \(f'(x) = \frac{{({x^2} + 5x + 4)(2x - 5) - ({x^2} - 5x + 4)(2x + 5)}}{{{{({x^2} + 5x + 4)}^2}}}\)     M1A1A1

\( = \frac{{10{x^2} - 40}}{{{{({x^2} + 5x + 4)}^2}}}\,\,\,\,\,\left( { = \frac{{10(x - 2)(x + 2)}}{{{{({x^2} + 5x + 4)}^2}}}} \right)\)     A1

\(f'(x) = 0 \Rightarrow x = \pm 2\)     M1

so the points under consideration are (–2, –9) and \(\left( {2, - \frac{1}{9}} \right)\)     A1A1

looking at the sign either side of the points (or attempt to find \(f''(x)\))     M1

e.g. if \(x = - {2^ - }\) then \((x - 2)(x + 2) > 0\) and if \(x = - {2^ + }\) then \((x - 2)(x + 2) < 0\),

therefore (–2, –9) is a maximum     A1

 

(ii)     e.g. if \(x = {2^ - }\) then \((x - 2)(x + 2) < 0\) and if \(x = {2^ + }\) then \((x - 2)(x + 2) > 0\),

therefore \(\left( {2, - \frac{1}{9}} \right)\) is a minimum     A1

Note: Candidates may find the minimum first.

 

[10 marks]

 

(d)

     A3

Note: Award A1 for each branch consistent with and including the features found in previous parts.

 

[3 marks]

 

(e)     one     A1

[1 mark]

Total [20 marks]

This was the most successfully answered question in part B, in particular parts (a), (b) and (c). In part (a) the horizontal asymptote was often missing (or x = 4, x = 1 given). Part (b) was well done. Use of the quotient rule was well done in part (c) and many simplified correctly. There was knowledge of max/min and how to justify their answer, usually with a sign diagram but also with the second derivative. A common misconception was that, as \( - 9 < - \frac{1}{9}\), the minimum is at (–2, –9). In part (d) many candidates were unable to sketch the graph consistent with the main features that they had determined before. Very few candidates answered part (e) correctly.