Given the graph y = (x2 − 9)(x + 3).

$$0 = (x-3)(x+3)(x+3)$$ $$0 = (x-3)(x+3)^2$$

$$x-3 = 0 \implies x = 3$$ $$x+3 = 0 \implies x = -3$$

The x-intercepts are $\boxed{(3,0) \text{ and } (-3,0)}$.

$$y = (-9)(3)$$ $$y = -27$$

The y-intercept is $\boxed{(0,-27)}$.

$$y = (x^2 - 9)(x + 3)$$ $$y = x^3 + 3x^2 - 9x - 27$$

$$\frac{dy}{dx} = \frac{d}{dx}(x^3 + 3x^2 - 9x - 27)$$ $$\frac{dy}{dx} = 3x^2 + 6x - 9$$

$$3x^2 + 6x - 9 = 0$$

Divide the entire equation by 3:

$$x^2 + 2x - 3 = 0$$

Factor the quadratic equation:

$$(x+3)(x-1) = 0$$

This gives two possible x-values:

$$x+3 = 0 \implies x = -3$$ $$x-1 = 0 \implies x = 1$$

For $x=-3$:

$$y = (-3)^3 + 3(-3)^2 - 9(-3) - 27$$ $$y = -27 + 3(9) + 27 - 27$$ $$y = -27 + 27 + 27 - 27$$ $$y = 0$$

So, one stationary point is $(-3,0)$.
For $x=1$:

$$y = (1)^3 + 3(1)^2 - 9(1) - 27$$ $$y = 1 + 3 - 9 - 27$$ $$y = 4 - 36$$ $$y = -32$$

So, the other stationary point is $(1,-32)$.
The stationary points are $\boxed{(-3,0) \text{ and } (1,-32)}$.

$$\frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 + 6x - 9)$$ $$\frac{d^2y}{dx^2} = 6x + 6$$

For $x=-3$:

$$\frac{d^2y}{dx^2}\Big|_{x=-3} = 6(-3) + 6 = -18 + 6 = -12$$

Since $\frac{d^2y}{dx^2} < 0$, the point $(-3,0)$ is a maximum point.
For $x=1$:

$$\frac{d^2y}{dx^2}\Big|_{x=1} = 6(1) + 6 = 6 + 6 = 12$$

Since $\frac{d^2y}{dx^2} > 0$, the point $(1,-32)$ is a minimum point.
So, $\boxed{(-3,0) \text{ is a maximum point while } (1,-32) \text{ is a minimum point}}$.