Problem Statement
Consider the rational function $f(x) = \frac{2x^2 + 3x - 5}{x - 2}$ for $x \neq 2$.
**(a)** [3 marks] Find the equations of the vertical asymptote and the oblique (slant) asymptote of the graph of $y = f(x)$.
**(b)** [5 marks] Find the coordinates of the two stationary points of $f$.
Verified Solution & Marking Scheme
Part (a) [A1]
Vertical asymptote occurs at $x = 2$.
Part (a) [M1]
Perform polynomial long division: $(2x^2 + 3x - 5) \div (x - 2) = 2x + 7 + \frac{9}{x - 2}$
Part (a) [A1]
As $x \to \pm\infty$, $\frac{9}{x-2} \to 0$. Oblique asymptote is $y = 2x + 7$.
Part (b) [M1]
Differentiate $f(x) = 2x + 7 + 9(x - 2)^{-1} \implies f'(x) = 2 - \frac{9}{(x - 2)^2}$
Part (b) [M1]
Set $f'(x) = 0 \implies \frac{9}{(x - 2)^2} = 2 \implies (x - 2)^2 = \frac{9}{2} \implies x - 2 = \pm\frac{3}{\sqrt{2}}$
Part (b) [A1]
$x_1 = 2 + \frac{3\sqrt{2}}{2}, \quad x_2 = 2 - \frac{3\sqrt{2}}{2}$
Part (b) [M1, A1]
Evaluate $y$-coordinates: $f(x_1) = 2(2 + \frac{3\sqrt{2}}{2}) + 7 + \frac{9}{3\sqrt{2}/2} = 11 + 3\sqrt{2} + 3\sqrt{2} = 11 + 6\sqrt{2}$. Coordinates: $(2 + \frac{3\sqrt{2}}{2}, 11 + 6\sqrt{2})$ and $(2 - \frac{3\sqrt{2}}{2}, 11 - 6\sqrt{2})$.