A curve C has equation $y = x + \frac{9}{x+1}$.

1. Find the first and second derivatives of $y$ with respect to $x$.

$y = x + 9(x+1)^{-1}$
$\frac{dy}{dx} = 1 - 9(x+1)^{-2}$
$\frac{d^2y}{dx^2} = 18(x+1)^{-3}$

2. Set up and solve the inequality for the curve to be concave upwards.

For a curve to be concave upwards, $\frac{d^2y}{dx^2} > 0$.
$\frac{18}{(x+1)^3} > 0$
Since $18 > 0$, we must have the denominator strictly positive:
$(x+1)^3 > 0$
$x + 1 > 0$
$x > -1$

1. Identify the equations of the asymptotes.

From the equation $y = x + \frac{9}{x+1}$:
As $x \to -1$, $\frac{9}{x+1} \to \pm \infty$. Thus, the vertical asymptote is $x = -1$.
As $x \to \pm \infty$, $\frac{9}{x+1} \to 0$. Thus, the oblique asymptote is $y = x$.

2. Find the coordinates of the stationary points.

Set $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$:
$1 - \frac{9}{(x+1)^2} = 0$
$(x+1)^2 = 9$
$x + 1 = \pm 3$
$x = 2 \quad \text{or} \quad x = -4$
When $x = 2$, $y = 2 + \frac{9}{3} = 5$. This gives a local minimum at $(2, 5)$.
When $x = -4$, $y = -4 + \frac{9}{-3} = -7$. This gives a local maximum at $(-4, -7)$.

3. Determine the intersections with the coordinate axes.

To find the $y$-intercept, set $x = 0$:
$y = 0 + \frac{9}{0+1} = 9$
The curve crosses the $y$-axis at $(0, 9)$.

To find $x$-intercepts, set $y = 0$:
$x + \frac{9}{x+1} = 0$
$x^2 + x + 9 = 0$
The discriminant is $1^2 - 4(1)(9) = -35 < 0$, so there are no real roots and therefore no $x$-intercepts.