(Q3) [Cambridge/H2 Maths 2007] 证明方程 $y = \dfrac{2x + 7}{x + 2}$ 可以写成 $y = A + \dfrac{B}{x + 2}$ 的形式，其中 $A$ 和 $B$ 是待求常数。由此说明将 $y = \dfrac{1}{x}$ 的图像变换为 $y = \dfrac{2x + 7}{x + 2}$ 的图像所经由的一系列变换步骤。

1. Express the rational function in the required form using algebraic manipulation.

$y = \dfrac{2x + 7}{x + 2}$

$y = \dfrac{2(x + 2) + 3}{x + 2}$

$y = 2 + \dfrac{3}{x + 2}$

Therefore, $A = 2$ and $B = 3$.

2. Identify the horizontal translation.

Translation by $2$ units in the negative $x$-direction.
This transforms $y = \dfrac{1}{x}$ to $y = \dfrac{1}{x+2}$.

3. Identify the vertical scaling.

Scaling parallel to the $y$-axis by a scale factor of $3$.
This transforms $y = \dfrac{1}{x+2}$ to $y = \dfrac{3}{x+2}$.

4. Identify the vertical translation.

Translation by $2$ units in the positive $y$-direction.
This transforms $y = \dfrac{3}{x+2}$ to $y = \dfrac{3}{x+2} + 2$.

1. Determine the equations of the vertical and horizontal asymptotes.

From the equation $y = \dfrac{2x + 7}{x + 2}$ (or equivalently $y = 2 + \dfrac{3}{x + 2}$ obtained from part (a)), the denominator is zero when $x + 2 = 0$, so the vertical asymptote is $x = -2$.

As $x \to \pm\infty$, $y \to 2$, so the horizontal asymptote is $y = 2$.

2. Calculate the coordinates of the points of intersection with the $x$- and $y$-axes.

To find the $y$-intercept, set $x = 0$:
$y = \dfrac{2(0) + 7}{0 + 2} = \dfrac{7}{2} = 3.5$
The intersection with the $y$-axis is $(0, 3.5)$.

To find the $x$-intercept, set $y = 0$:
$0 = \dfrac{2x + 7}{x + 2}$
$2x + 7 = 0 \implies x = -3.5$
The intersection with the $x$-axis is $(-3.5, 0)$.

3. Describe and draw the sketch of the function.

Draw dashed lines for the asymptotes $x = -2$ and $y = 2$. Plot the intercepts $(-3.5, 0)$ and $(0, 3.5)$. Draw a smooth curve consisting of two separate branches typical of a hyperbola $y = A + \frac{B}{x+C}$ where $B > 0$: one branch in the bottom-left passing through $(-3.5, 0)$ and bound by the asymptotes, and one branch in the top-right passing through $(0, 3.5)$ and bound by the asymptotes.