The diagram below shows part of the graph of $y = \frac{x^2 + 2x + 5}{x + 3}$. Find the coordinates of the minimum point $P$.

1. Find the derivative $\frac{dy}{dx}$ using the quotient rule.

Let $u = x^2 + 2x + 5$ and $v = x + 3$.
Then $\frac{du}{dx} = 2x + 2$ and $\frac{dv}{dx} = 1$.

$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$
$\frac{dy}{dx} = \frac{(x + 3)(2x + 2) - (x^2 + 2x + 5)(1)}{(x + 3)^2}$
$\frac{dy}{dx} = \frac{2x^2 + 2x + 6x + 6 - x^2 - 2x - 5}{(x + 3)^2}$
$\frac{dy}{dx} = \frac{x^2 + 6x + 1}{(x + 3)^2}$

2. Set the derivative to zero to find the $x$-coordinates of the stationary points.

For a minimum point, $\frac{dy}{dx} = 0$.
$\frac{x^2 + 6x + 1}{(x + 3)^2} = 0$
$x^2 + 6x + 1 = 0$

3. Solve the quadratic equation for $x$.

Using the quadratic formula:
$x = \frac{-6 \pm \sqrt{6^2 - 4(1)(1)}}{2(1)}$
$x = \frac{-6 \pm \sqrt{32}}{2}$
$x = \frac{-6 \pm 4\sqrt{2}}{2}$
$x = -3 \pm 2\sqrt{2}$

4. Determine the correct $x$-coordinate for point $P$ based on the graph.

The graph has a vertical asymptote at $x = -3$. The point $P$ lies on the branch of the curve that crosses the positive $y$-axis, which means $x > -3$.
Therefore, we reject $x = -3 - 2\sqrt{2}$ and accept $x = -3 + 2\sqrt{2}$.

5. Find the corresponding $y$-coordinate for point $P$.

Substitute $x = -3 + 2\sqrt{2}$ into the original equation:
$y = \frac{(-3 + 2\sqrt{2})^2 + 2(-3 + 2\sqrt{2}) + 5}{(-3 + 2\sqrt{2}) + 3}$
$y = \frac{9 - 12\sqrt{2} + 8 - 6 + 4\sqrt{2} + 5}{2\sqrt{2}}$
$y = \frac{16 - 8\sqrt{2}}{2\sqrt{2}}$
$y = \frac{8 - 4\sqrt{2}}{\sqrt{2}}$
$y = \frac{8\sqrt{2} - 8}{2} = 4\sqrt{2} - 4$

The coordinates of $P$ are $(-3 + 2\sqrt{2}, -4 + 4\sqrt{2})$.