Find the equations of the horizontal and vertical asymptotes of the curve \(y = f(x)\).

(i)     Find \(f'(x)\).

(ii)     Show that the curve has exactly one point where its tangent is horizontal.

(iii)     Find the coordinates of this point.

Find the equation of \({L_1}\), the normal to the curve at the point where it crosses the y-axis.

Find the equation of the line \({L_2}\).

\(x \to  - \infty  \Rightarrow y \to  - \frac{1}{2}\) so \(y =  - \frac{1}{2}\) is an asymptote     (M1)A1

\({{\text{e}}^x} - 2 = 0 \Rightarrow x = \ln 2\) so \(x = \ln 2{\text{ }}( = 0.693)\) is an asymptote     (M1)A1

[4 marks]

(i)     \(f'(x) = \frac{{2\left( {{{\text{e}}^x} - 2} \right){{\text{e}}^{2x}} - \left( {{{\text{e}}^{2x}} + 1} \right){{\text{e}}^x}}}{{{{\left( {{{\text{e}}^x} - 2} \right)}^2}}}\)     M1A1

          \( = \frac{{{{\text{e}}^{3x}} - 4{{\text{e}}^{2x}} - {{\text{e}}^x}}}{{{{\left( {{{\text{e}}^x} - 2} \right)}^2}}}\)

(ii)     \(f'(x) = 0\) when \({{\text{e}}^{3x}} - 4{{\text{e}}^{2x}} - {{\text{e}}^x} = 0\)     M1

          \({{\text{e}}^x}\left( {{{\text{e}}^{2x}} - 4{{\text{e}}^x} - 1} \right) = 0\)

          \({{\text{e}}^x} = 0,{\text{ }}{{\text{e}}^x} =  - 0.236,{\text{ }}{{\text{e}}^x} = 4.24{\text{ }}({\text{or }}{{\text{e}}^x} = 2 \pm \sqrt 5 )\)     A1A1

Note:     Award A1 for zero, A1 for other two solutions.

     Accept any answers which show a zero, a negative and a positive.

          as \({{\text{e}}^x} > 0\) exactly one solution     R1

Note:     Do not award marks for purely graphical solution.

(iii)     (1.44, 8.47)     A1A1

[8 marks]

\(f'(0) =  - 4\)     (A1)

so gradient of normal is \(\frac{1}{4}\)     (M1)

\(f(0) =  - 2\)     (A1)

so equation of \({L_1}\) is \(y = \frac{1}{4}x - 2\)     A1

[4 marks]

\(f'(x) = \frac{1}{4}\)     M1

so \(x = 1.46\)     (M1)A1

\(f(1.46) = 8.47\)     (A1)

equation of \({L_2}\) is \(y - 8.47 = \frac{1}{4}(x - 1.46)\)     A1

(or \(y = \frac{1}{4}x + 8.11\))

[5 marks]