Find the equation of the normal to the curve $y = \frac{2x-6}{x-2}$ at the point where the curve meets the y-axis. Determine, with explanation, whether $y$ is an increasing or decreasing function for $x > 2$. A curve is such that $\dfrac{d^{2}y}{dx^{2}} = (3x+1)^{2}$ and the point $Q(2,36)$ lies on the curve. The gradient of the curve at $Q$ is $45$. Find the equation of the curve.

1. Find the coordinates of the point where the curve meets the y-axis.

When $x = 0$,
$y = \frac{2(0)-6}{0-2} = \frac{-6}{-2} = 3$
The point is $(0, 3)$.

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

Let $u = 2x - 6$ and $v = x - 2$.
$\frac{du}{dx} = 2$ and $\frac{dv}{dx} = 1$.
$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$
$\frac{dy}{dx} = \frac{(x-2)(2) - (2x-6)(1)}{(x-2)^2}$
$\frac{dy}{dx} = \frac{2x - 4 - 2x + 6}{(x-2)^2} = \frac{2}{(x-2)^2}$

3. Calculate the gradient of the tangent at $x = 0$.

At $x = 0$,
$\frac{dy}{dx} = \frac{2}{(0-2)^2} = \frac{2}{4} = \frac{1}{2}$

4. Find the gradient of the normal and the equation of the normal.

Gradient of normal, $m_n = -\frac{1}{\frac{1}{2}} = -2$.
Equation of the normal at $(0, 3)$:
$y - 3 = -2(x - 0)$
$y = -2x + 3$

1. Find the first derivative of $y$ with respect to $x$.

$y = \frac{2x-6}{x-2}$
Using the quotient rule:
$\frac{dy}{dx} = \frac{2(x-2) - (2x-6)(1)}{(x-2)^2}$
$\frac{dy}{dx} = \frac{2x - 4 - 2x + 6}{(x-2)^2}$
$\frac{dy}{dx} = \frac{2}{(x-2)^2}$

2. Determine the sign of the derivative for $x > 2$ and conclude.

For $x > 2$, $(x-2)^2 > 0$.
Therefore, $\frac{dy}{dx} = \frac{2}{(x-2)^2} > 0$.
Since the derivative is positive, $y$ is an increasing function.

1. Integrate the second derivative to find the expression for the first derivative, $\frac{dy}{dx}$.

$\frac{dy}{dx} = \int (3x+1)^2 \, dx$
$\frac{dy}{dx} = \frac{(3x+1)^3}{3(3)} + C$
$\frac{dy}{dx} = \frac{(3x+1)^3}{9} + C$

2. Use the given gradient at $Q(2,36)$ to find the constant of integration $C$.

At $x = 2$, $\frac{dy}{dx} = 45$:
$45 = \frac{(3(2)+1)^3}{9} + C$
$45 = \frac{7^3}{9} + C$
$45 = \frac{343}{9} + C$
$C = 45 - \frac{343}{9} = \frac{405 - 343}{9} = \frac{62}{9}$
So, $\frac{dy}{dx} = \frac{(3x+1)^3}{9} + \frac{62}{9}$

3. Integrate $\frac{dy}{dx}$ to find the expression for $y$.

$y = \int \left( \frac{(3x+1)^3}{9} + \frac{62}{9} \right) \, dx$
$y = \frac{(3x+1)^4}{9(4)(3)} + \frac{62}{9}x + D$
$y = \frac{(3x+1)^4}{108} + \frac{62}{9}x + D$

4. Use the coordinates of point $Q$ to find the second constant of integration $D$.

At $x = 2$, $y = 36$:
$36 = \frac{(3(2)+1)^4}{108} + \frac{62}{9}(2) + D$
$36 = \frac{7^4}{108} + \frac{124}{9} + D$
$36 = \frac{2401}{108} + \frac{1488}{108} + D$
$36 = \frac{3889}{108} + D$
$D = 36 - \frac{3889}{108} = \frac{3888 - 3889}{108} = -\frac{1}{108}$

5. State the final equation of the curve.

$y = \frac{(3x+1)^4}{108} + \frac{62}{9}x - \frac{1}{108}$