曲线 $C$ 的方程为 $y = \dfrac{x-3}{(x-2)(x+5)}$。画出 $C$ 的草图，并给出渐近线方程、驻点坐标以及 $C$ 与坐标轴的交点坐标。 描述将 $C$ 映射到 $y = \dfrac{2x-3}{(x-1)(2x+5)}$ 的图像上的一系列变换。

1. Find the coordinates of the points where the curve crosses the axes.

To find the $y$-intercept, substitute $x = 0$ into the equation:
$y = \dfrac{0-3}{(0-2)(0+5)} = \dfrac{-3}{-10} = \dfrac{3}{10}$
The $y$-intercept is $\left(0, \dfrac{3}{10}\right)$.

To find the $x$-intercept, substitute $y = 0$:
$\dfrac{x-3}{(x-2)(x+5)} = 0 \implies x - 3 = 0 \implies x = 3$
The $x$-intercept is $(3, 0)$.

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

The function is undefined when the denominator is zero:
$(x-2)(x+5) = 0 \implies x = 2 \text{ or } x = -5$
The equations of the vertical asymptotes are $x = 2$ and $x = -5$.

As $x \to \pm\infty$, the denominator grows proportionally faster than the numerator (degree 2 vs degree 1), so $y \to 0$.
The equation of the horizontal asymptote is $y = 0$.

3. Find the coordinates of the stationary points.

Using a graphing calculator to find the maximum and minimum turning points of $y = \dfrac{x-3}{(x-2)(x+5)}$:

• Local minimum point $\approx (0.172, 0.299)$
• Local maximum point $\approx (5.83, 0.0682)$

(Alternatively, by setting $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$, exact coordinates are $\left(3 - 2\sqrt{2}, \dfrac{9 + 4\sqrt{2}}{49}\right)$ and $\left(3 + 2\sqrt{2}, \dfrac{9 - 4\sqrt{2}}{49}\right)$, though 3 s.f. is standard and sufficient via GC unless exact forms are explicitly requested.)

4. Sketch the curve, incorporating all identified features.

Draw the axes and sketch the curve ensuring the following are clearly labeled:

• Asymptotes: Dashed lines at $x = -5$ and $x = 2$, and label the $x$-axis as $y = 0$.
• Intercepts: Points labeled at $(0, 0.3)$ and $(3, 0)$.
• Stationary points: A local minimum marked at $(0.172, 0.299)$ and a local maximum marked at $(5.83, 0.0682)$.
• The curve has three branches: a monotonically increasing branch entirely below the $x$-axis for $x < -5$, a U-shaped branch in $-5 < x < 2$ opening upwards with the local minimum, and a branch for $x > 2$ coming from negative infinity, crossing the $x$-axis, hitting the local maximum, and gradually falling towards the $x$-axis.

1. Express the new equation in terms of the original function $\mathrm{f}(x)$.

Let $\mathrm{f}(x) = \dfrac{x-3}{(x-2)(x+5)}$.

Substitute $x$ with $2x$:
$\mathrm{f}(2x) = \dfrac{2x-3}{(2x-2)(2x+5)} = \dfrac{2x-3}{2(x-1)(2x+5)}$

Multiply by 2:
$2\mathrm{f}(2x) = 2 \left[ \dfrac{2x-3}{2(x-1)(2x+5)} \right] = \dfrac{2x-3}{(x-1)(2x+5)}$

The new equation is $y = 2\mathrm{f}(2x)$.

2. Describe the transformation associated with $f(2x)$.

The transformation from $\mathrm{f}(x)$ to $\mathrm{f}(2x)$ is a scaling (or stretch) parallel to the $x$-axis by a scale factor of $\dfrac{1}{2}$.

3. Describe the transformation associated with the multiplier $2$.

The transformation from $\mathrm{f}(2x)$ to $2\mathrm{f}(2x)$ is a scaling (or stretch) parallel to the $y$-axis by a scale factor of $2$.

Since the transformations operate on different axes, the sequence of these transformations can be done in any order.