Answer: \(\dfrac{dy}{dx}=\dfrac{-\bigl((x+2)^2+5\bigr)}{(x^2-9)^2}\), so \(\dfrac{dy}{dx}\lt0\) at all points on \(C\).

The asymptotes are \(x=-3\), \(x=3\), and \(y=0\).

The curve crosses the axes at \((-2,0)\) and \(\left(0,-\dfrac{2}{9}\right)\).

Differentiate \(y=\dfrac{x+2}{x^2-9}\) using the quotient rule:

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

Simplifying the numerator gives

\(x^2-9-2x^2-4x=-x^2-4x-9=-\bigl((x+2)^2+5\bigr).\)

Hence

\(\frac{dy}{dx}=\frac{-\bigl((x+2)^2+5\bigr)}{(x^2-9)^2}.\)

The denominator is positive wherever the curve is defined, and \((x+2)^2+5\gt0\), so \(\dfrac{dy}{dx}\lt0\) at every point on \(C\).

The vertical asymptotes occur where \(x^2-9=0\), so \(x=-3\) and \(x=3\). Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is \(y=0\).

For the intercepts, set \(y=0\). This gives \(x+2=0\), so the \(x\)-intercept is \((-2,0)\). Setting \(x=0\) gives \(y=\dfrac{2}{-9}=-\dfrac{2}{9}\), so the \(y\)-intercept is \(\left(0,-\dfrac{2}{9}\right)\).

The sketch has three decreasing branches separated by the vertical asymptotes \(x=-3\) and \(x=3\). The left branch approaches \(y=0\) from below as \(x\to-\infty\), the middle branch crosses the axes, and the right branch approaches \(y=0\) from above as \(x\to\infty\).