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

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

$y = x + 9(x+1)^{-1}$

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

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

2. Apply the condition for the curve to concave upwards and solve for $x$.

For a curve to concave upwards, $\frac{d^2y}{dx^2} > 0$.

$\frac{18}{(x+1)^3} > 0$

Since $18 > 0$, we must have $(x+1)^3 > 0$.

$x + 1 > 0$

$x > -1$

1. Identify the vertical and oblique asymptotes.

The term $\frac{9}{x+1}$ is undefined when $x = -1$, which gives the vertical asymptote $x = -1$.
As $x \to \pm\infty$, $\frac{9}{x+1} \to 0$, so the curve approaches the oblique asymptote $y = x$.

2. Find the stationary points and axial intercepts.

Setting $\frac{dy}{dx} = 1 - \frac{9}{(x+1)^2} = 0$ gives $(x+1)^2 = 9 \implies x = 2 \text{ or } x = -4$.
Substituting these into the curve's equation gives the turning points $(2, 5)$ and $(-4, -7)$.
When $x = 0$, $y = 0 + \frac{9}{1} = 9$, making the $y$-intercept $(0, 9)$.
Setting $y = 0$ yields $x^2 + x + 9 = 0$. The discriminant $1^2 - 4(1)(9) < 0$ indicates no real roots, so there are no $x$-intercepts.

3. Sketch the graph.

Draw the coordinate axes and the asymptotes $x = -1$ and $y = x$ with dashed lines.
Plot the local maximum point $(-4, -7)$ and sketch the left branch concave downwards bounded by the asymptotes.
Plot the local minimum point $(2, 5)$ and the $y$-intercept $(0, 9)$, and sketch the right branch concave upwards, also bounded by the asymptotes.