在同一坐标系中，画出 $y = \frac{1}{x - 2}$ 和 $y = 4x - 2$ 的函数图像，并清晰地标出渐近线方程以及与坐标轴交点的坐标。 由此或通过其他方法，解不等式 $\frac{1}{x - 2} \le 4x - 2$。

1. Find the asymptotes and axial intercepts of the rational function $y = \frac{1}{x - 2}$.

For $y = \frac{1}{x - 2}$:

• The denominator cannot be zero, giving a vertical asymptote at $x = 2$.
• As $x \to \pm\infty$, $y \to 0$, giving a horizontal asymptote at $y = 0$ (the $x$-axis).
• When $x = 0$, $y = \frac{1}{0 - 2} = -\frac{1}{2}$. The $y$-intercept is $(0, -\frac{1}{2})$.
• There is no $x$-intercept since the numerator is a non-zero constant ($1 \neq 0$).

2. Find the axial intercepts of the linear function $y = 4x - 2$.

For $y = 4x - 2$:

• When $x = 0$, $y = 4(0) - 2 = -2$. The $y$-intercept is $(0, -2)$.
• When $y = 0$, $4x - 2 = 0 \implies x = \frac{1}{2}$. The $x$-intercept is $(\frac{1}{2}, 0)$.

3. Sketch both graphs on the same axes.

On a single set of axes:

• Draw a dashed vertical line labelled $x = 2$.
• Draw the two branches of the hyperbola for $y = \frac{1}{x - 2}$, labelling the intercept $(0, -\frac{1}{2})$.
• Draw a straight line for $y = 4x - 2$, indicating the intercepts $(0, -2)$ and $(\frac{1}{2}, 0)$.

1. Find the $x$-coordinates of the points of intersection between the curve and the line.

Equating the two expressions:
$\frac{1}{x - 2} = 4x - 2$
$1 = (4x - 2)(x - 2)$
$1 = 4x^2 - 10x + 4$
$4x^2 - 10x + 3 = 0$
Using the quadratic formula:
$x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(4)(3)}}{2(4)}$
$x = \frac{10 \pm \sqrt{52}}{8} = \frac{5 \pm \sqrt{13}}{4}$

2. Use the graphs from part (i) to deduce the intervals where the curve is below or equal to the line.

The inequality $\frac{1}{x - 2} \le 4x - 2$ corresponds to the regions where the curve $y = \frac{1}{x - 2}$ lies below or exactly on the line $y = 4x - 2$.

From the graphs sketched in (i), taking into account the relative positions of the curve and the line alongside the vertical asymptote at $x = 2$:

• For $x < 2$, the curve is below the line starting from the first intersection point $x = \frac{5 - \sqrt{13}}{4}$ up to the vertical asymptote (where it tends towards $-\infty$). Note that $x$ cannot be exactly 2.
• For $x > 2$, the curve emerges from $+\infty$ and drops below the line starting from the second intersection point $x = \frac{5 + \sqrt{13}}{4}$ onwards.

Thus, the combined solution is:
$\frac{5 - \sqrt{13}}{4} \le x < 2 \quad \text{or} \quad x \ge \frac{5 + \sqrt{13}}{4}$