1. **Problem statement:** Given the function $$f(x) = \frac{x^2 + 12}{x - t}$$ with a vertical asymptote at $$x = 2$$, a non-vertical asymptote $$l$$, and the graph crossing the y-axis at point $$P$$. Points $$Q$$ and $$R$$ are stationary points. We need to find: (a) The value of $$t$$. (b) The equation of the non-vertical asymptote $$l$$. (c) The coordinates of point $$P$$ where the graph crosses the y-axis. (d) The coordinates of the stationary points using calculus. --- 2. **(a) Find the value of $$t$$:** The vertical asymptote occurs where the denominator is zero, so: $$x - t = 0 \implies x = t$$ Given the vertical asymptote is at $$x = 2$$, therefore: $$t = 2$$ --- 3. **(b) Find the non-vertical asymptote $$l$$:** For large $$|x|$$, divide numerator by denominator using polynomial division: Divide $$x^2 + 12$$ by $$x - 2$$: $$ \begin{aligned} &x^2 \div x = x \\ &x \times (x - 2) = x^2 - 2x \\ &\text{Subtract: } (x^2 + 12) - (x^2 - 2x) = 2x + 12 \\ &2x \div x = 2 \\ &2 \times (x - 2) = 2x - 4 \\ &\text{Subtract: } (2x + 12) - (2x - 4) = 16 \end{aligned} $$ So, $$ \frac{x^2 + 12}{x - 2} = x + 2 + \frac{16}{x - 2} $$ As $$x \to \pm \infty$$, $$\frac{16}{x - 2} \to 0$$, so the non-vertical asymptote is: $$ \boxed{y = x + 2} $$ --- 4. **(c) Find coordinates of point $$P$$ where the graph crosses the y-axis:** At the y-axis, $$x = 0$$, so: $$ f(0) = \frac{0^2 + 12}{0 - 2} = \frac{12}{-2} = -6 $$ Therefore, $$ P = (0, -6) $$ --- 5. **(d) Find stationary points $$Q$$ and $$R$$ using calculus:** Stationary points occur where $$f'(x) = 0$$. Given: $$ f(x) = \frac{x^2 + 12}{x - 2} $$ Use quotient rule: $$ f'(x) = \frac{(2x)(x - 2) - (x^2 + 12)(1)}{(x - 2)^2} $$ Simplify numerator: $$ 2x(x - 2) - (x^2 + 12) = 2x^2 - 4x - x^2 - 12 = x^2 - 4x - 12 $$ Set numerator equal to zero for stationary points: $$ x^2 - 4x - 12 = 0 $$ Solve quadratic: $$ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times (-12)}}{2} = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2} = \frac{4 \pm 8}{2} $$ So, $$ x_1 = \frac{4 + 8}{2} = 6, \quad x_2 = \frac{4 - 8}{2} = -2 $$ Find corresponding $$y$$ values: $$ f(6) = \frac{6^2 + 12}{6 - 2} = \frac{36 + 12}{4} = \frac{48}{4} = 12 $$ $$ f(-2) = \frac{(-2)^2 + 12}{-2 - 2} = \frac{4 + 12}{-4} = \frac{16}{-4} = -4 $$ Therefore, stationary points are: $$ Q = (6, 12), \quad R = (-2, -4) $$