函数 $y = \frac{ax^2 + bx + c}{x+d}$ 的图像有一条方程为 $x = -1$ 的垂直渐近线和一条方程为 $y = 2x + 2$ 的斜渐近线。已知 $c \ne 2$。

1. Find the value of $d$ using the given vertical asymptote.

A vertical asymptote occurs when the denominator evaluates to zero (and the numerator is non-zero).
Given the vertical asymptote is $x = -1$, the denominator $x + d$ must be zero when $x = -1$.
$-1 + d = 0 \implies d = 1$

2. Express the rational function in quotient-remainder form to identify its oblique asymptote.

Substitute $d = 1$ into the function equation:
$y = \frac{ax^2 + bx + c}{x + 1}$
By performing polynomial long division or expressing the numerator in terms of the denominator:
$ax^2 + bx + c = ax(x + 1) + (b - a)(x + 1) + c - (b - a)$
$y = ax + (b - a) + \frac{c - b + a}{x + 1}$
For large values of $|x|$, the fractional term $\frac{c - b + a}{x + 1}$ approaches $0$.
Therefore, the equation of the oblique asymptote is $y = ax + (b - a)$.

3. Compare coefficients with the given oblique asymptote to show $a = 2$ and $b = 4$.

The given oblique asymptote is $y = 2x + 2$. Comparing this with $y = ax + (b - a)$:
Comparing the coefficient of $x$:
$a = 2$
Comparing the constant term:
$b - a = 2$
$b - 2 = 2 \implies b = 4$

1. Substitute the values of $a$, $b$, and $d$ from part (a) into the function and differentiate with respect to $x$.

From part (a), $a = 2$, $b = 4$ and $d = 1$, so the equation of the curve is
$y = \frac{2x^2 + 4x + c}{x+1}$
Differentiating using the quotient rule:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(4x+4)(x+1) - (2x^2+4x+c)(1)}{(x+1)^2}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4x^2 + 8x + 4 - 2x^2 - 4x - c}{(x+1)^2}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x^2 + 4x + 4 - c}{(x+1)^2}$

2. Apply the condition for no stationary points by considering the discriminant of the numerator.

For stationary points, $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$, which requires the numerator to be zero:
$2x^2 + 4x + (4 - c) = 0$
For the graph to have no stationary points, this quadratic equation must have no real roots (since the denominator $(x+1)^2 \neq 0$ for $x \neq -1$). Thus, the discriminant must be strictly less than zero:
$\Delta < 0$
$(4)^2 - 4(2)(4 - c) < 0$
$16 - 32 + 8c < 0$
$8c < 16$
$c < 2$