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]