曲线 $C$ 的方程为 $y = \dfrac{x^2 + (\lambda - 1)x}{x - 1}, \quad x \ne 1,$ 其中 $\lambda$ 为常数。求

1. Find the vertical asymptote by setting the denominator to zero

The function is undefined when the denominator is zero:
$x - 1 = 0 \implies x = 1$
Thus, the vertical asymptote is $x = 1$.

2. Rewrite the function using algebraic manipulation or long division to find the oblique asymptote

By algebraic manipulation (or polynomial long division):
$y = \frac{x^2 - x + \lambda x}{x - 1}$
$y = \frac{x(x - 1) + \lambda(x - 1) + \lambda}{x - 1}$
$y = x + \lambda + \frac{\lambda}{x - 1}$

3. Identify the oblique asymptote

As $x \to \pm\infty$, the fractional term $\frac{\lambda}{x - 1} \to 0$.
Therefore, the oblique (slant) asymptote is $y = x + \lambda$.

1. Rewrite the equation of the curve to simplify differentiation and find the first derivative.

Using algebraic manipulation (as typically done to find asymptotes in part (i)):
$y = \frac{x^2 + (\lambda - 1)x}{x - 1} = \frac{x(x - 1) + \lambda x}{x - 1} = x + \lambda + \frac{\lambda}{x - 1}$
Differentiating with respect to $x$:
$\frac{dy}{dx} = 1 - \frac{\lambda}{(x - 1)^2}$

2. Equate the derivative to zero to find the conditions for stationary points.

At stationary points, $\frac{dy}{dx} = 0$:
$1 - \frac{\lambda}{(x - 1)^2} = 0$
$(x - 1)^2 = \lambda$

3. Solve for $x$ and establish the condition for two distinct real roots.

Taking the square root on both sides yields the $x$-coordinates of the stationary points:
$x = 1 \pm \sqrt{\lambda}$
For the curve to have two distinct real stationary points, we must be able to evaluate the square root to two distinct values, which requires:
$\lambda > 0$

4. Apply the condition that both stationary points occur for $x > 0$ to find the final range of $\lambda$.

The question requires both stationary points to be strictly positive ($x > 0$). This means the smaller root must also be strictly greater than zero:
$1 - \sqrt{\lambda} > 0$
$\sqrt{\lambda} < 1$
$\lambda < 1$
Combining this with the condition $\lambda > 0$, the range of values for $\lambda$ is:
$0 < \lambda < 1$

1. Find the intercepts with the axes.

Set $y = 0$:
$x^2 + (\lambda - 1)x = 0$
$x(x + \lambda - 1) = 0$
$x = 0 \quad \text{or} \quad x = 1 - \lambda$
From (ii), the range of values is $0 < \lambda < 1$, so $1 - \lambda$ is positive and less than $1$.
The graph cuts the $x$-axis at $(0, 0)$ and $(1 - \lambda, 0)$.

2. Recall asymptotes and analyse the shape using the range of $\lambda$.

From (i), the asymptotes are $x = 1$ and $y = x + \lambda$.
Since $0 < \lambda < 1$, the oblique asymptote $y = x + \lambda$ crosses the $y$-axis at a positive value $\lambda$.
As $x \to \pm\infty$, the curve approaches $y = x + \lambda$.
For $x < 1$, $y < x + \lambda$ and it passes through $(0,0)$ and $(1-\lambda,0)$. This confirms a local maximum between $x=0$ and $x=1-\lambda$.
For $x > 1$, $y > x + \lambda$ with a local minimum in the first quadrant.

3. Sketch the graph.

Draw the vertical line $x=1$ and the slanted line $y=x+\lambda$.
Plot the $x$-intercepts at $(0,0)$ and $(1-\lambda, 0)$.
Draw the two curve branches, ensuring the stationary points for $x > 0$ are visibly in the correct relative positions.

1. Find the expression for the derivative $f'(x)$.

Using the quotient rule on $f(x) = \dfrac{x^2 + (\lambda - 1)x}{x - 1}$:

$f'(x) = \dfrac{(2x + \lambda - 1)(x - 1) - (x^2 + (\lambda - 1) x)(1)}{(x - 1)^2}$
$f'(x) = \dfrac{2x^2 - 2x + \lambda x - \lambda - x + 1 - x^2 - \lambda x + x}{(x - 1)^2}$
$f'(x) = \dfrac{x^2 - 2x + 1 - \lambda}{(x - 1)^2} = \dfrac{(x - 1)^2 - \lambda}{(x - 1)^2} = 1 - \dfrac{\lambda}{(x - 1)^2}$

2. Determine the equations of the asymptotes.

For the graph of $y = 1 - \dfrac{\lambda}{(x - 1)^2}$:

• As $x \to 1$, the denominator approaches $0$, so $y \to -\infty$ (since $\lambda > 0$). Hence, the vertical asymptote is $x = 1$.
• As $x \to \pm\infty$, $\dfrac{\lambda}{(x - 1)^2} \to 0$, so $y \to 1$. Hence, the horizontal asymptote is $y = 1$.

3. Find the coordinates of the points where the graph cuts the x-axis.

Set $f'(x) = 0$ to find the x-intercepts:
$1 - \dfrac{\lambda}{(x - 1)^2} = 0$
$(x - 1)^2 = \lambda$
$x - 1 = \pm\sqrt{\lambda} \implies x = 1 \pm \sqrt{\lambda}$
The coordinates where the graph cuts the x-axis are $(1 - \sqrt{\lambda}, 0)$ and $(1 + \sqrt{\lambda}, 0)$.

4. Determine the graph's key features to complete the sketch.

From part (ii), we know the required range is $0 < \lambda < 1$.
Evaluating the y-intercept at $x = 0$:
$y = 1 - \dfrac{\lambda}{(0 - 1)^2} = 1 - \lambda$
Because $0 < \lambda < 1$, the y-intercept is strictly positive.
Both x-intercepts are also positive since $1 - \sqrt{\lambda} > 0$.
This establishes the positioning of the two downward-opening rational branches relative to the origin.