Graphs and transformations: EJC 2024 Promo Q12

(a)

When $x = 0$: $$\begin{aligned} y &= -2 + \frac{-6k}{-2k} = -2 + 3 = 1. \end{aligned}$$ When $y = 0$: $$\begin{aligned} (x - 2)\left(x^2 - 2k\right) + 3k(x - 2) &= 0\\ (x - 2)\left(x^2 - 2k + 3k\right) &= 0\\ \Rightarrow (x - 2)\left(x^2 + k\right) &= 0\\ \Rightarrow x &= 2 \end{aligned}$$ The coordinates are $(0, 1)$ and $(2, 0)$.

(b)

As $x \to \infty$, $y \to x - 2$

Vertical asymptotes: $x=\sqrt{2k}$ and $x=-\sqrt{2k}$

Oblique asymptote: $y=x-2$

(c)

$$y=x-2+\frac{3(x-2)}{x^2-2}$$ $$\begin{aligned} \frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= 1 + 3\left[\frac{\left(x^2 - 2\right) - (x - 2)(2x)}{\left(x^2 - 2\right)^2}\right]\\ &= 1 + \frac{3\left(-x^2 + 4x - 2\right)}{\left(x^2 - 2\right)^2}. \end{aligned}$$ At turning points, $\dfrac{\,\mathrm{d} y}{\,\mathrm{d} x} = 0$.

Hence we have $$\begin{aligned} \left(x^2 - 2\right)^2 + 3\left(-x^2 + 4x - 2\right) &= 0\\ x^4 - 4x^2 + 4 - 3x^2 + 12x - 6 &= 0\\ x^4 - 7x^2 + 12x - 2 &= 0. \quad\text{((shown))} \end{aligned}$$ Using GC, $x = -3.29$ or $x = 0.187$.

(d)