A) Find the co-ordinates of the point at which x=2. b) Find the equation of the normal to the curve at the point in (a) above in double intercept form.

$$y = (2)^3 - 6(2)^2 + 9(2) + 5$$ $$y = 8 - 6(4) + 18 + 5$$ $$y = 8 - 24 + 18 + 5$$ $$y = -16 + 18 + 5$$ $$y = 2 + 5$$ $$y = 7$$

The coordinates of the point are $\boxed{(2, 7)}$.

$$y = x^3 - 6x^2 + 9x + 5$$ $$\frac{dy}{dx} = 3x^2 - 12x + 9$$

$$m_t = 3(2)^2 - 12(2) + 9$$ $$m_t = 3(4) - 24 + 9$$ $$m_t = 12 - 24 + 9$$ $$m_t = -12 + 9$$ $$m_t = -3$$

The normal is perpendicular to the tangent, so its gradient $m_n$ is the negative reciprocal of $m_t$.

$$m_n = -\frac{1}{m_t} = -\frac{1}{-3} = \frac{1}{3}$$

Using the point-slope form $y - y_1 = m_n(x - x_1)$:

$$y - 7 = \frac{1}{3}(x - 2)$$

Multiply by 3 to clear the fraction:

$$3(y - 7) = x - 2$$ $$3y - 21 = x - 2$$

Rearrange into the form $Ax + By = C$:

$$x - 3y = -19$$

Divide the entire equation by $-19$:

$$\frac{x}{-19} - \frac{3y}{-19} = \frac{-19}{-19}$$ $$\frac{x}{-19} + \frac{3y}{19} = 1$$

This can also be written as:

$$\frac{x}{-19} + \frac{y}{19/3} = 1$$

The equation of the normal in double intercept form is $\boxed{\frac{x}{-19} + \frac{y}{19/3} = 1}$.

$$\frac{dy}{dx} = 3x^2 - 12x + 9$$ $$3x^2 - 12x + 9 = 0$$

Divide by 3:

$$x^2 - 4x + 3 = 0$$

Factor the quadratic equation:

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

So, the x-coordinates of the stationary points are $x = 1$ and $x = 3$.

For $x = 1$:

$$y = (1)^3 - 6(1)^2 + 9(1) + 5$$ $$y = 1 - 6 + 9 + 5$$ $$y = -5 + 9 + 5$$ $$y = 4 + 5$$ $$y = 9$$

The first stationary point is $(1, 9)$.

For $x = 3$:

$$y = (3)^3 - 6(3)^2 + 9(3) + 5$$ $$y = 27 - 6(9) + 27 + 5$$ $$y = 27 - 54 + 27 + 5$$ $$y = -27 + 27 + 5$$ $$y = 5$$

The second stationary point is $(3, 5)$.

Find the second derivative $\frac{d^2y}{dx^2}$:

$$\frac{dy}{dx} = 3x^2 - 12x + 9$$ $$\frac{d^2y}{dx^2} = 6x - 12$$

Evaluate $\frac{d^2y}{dx^2}$ at each stationary point:
For $x = 1$:

$$\frac{d^2y}{dx^2} = 6(1) - 12 = 6 - 12 = -6$$

Since $\frac{d^2y}{dx^2} < 0$, the point $(1, 9)$ is a local maximum.

For $x = 3$:

$$\frac{d^2y}{dx^2} = 6(3) - 12 = 18 - 12 = 6$$

Since $\frac{d^2y}{dx^2} > 0$, the point $(3, 5)$ is a local minimum.

The stationary points are:
$\boxed{(1, 9) \text{ which is a local maximum}}$
$\boxed{(3, 5) \text{ which is a local minimum}}$