Slope of a Curve of Given Parametric Equations

$x = t^2 + 2t$

$dx = (2t + 2) \, dt$
 

$y = 2t^3 - 6t$

$dy = (6t^2 - 6) \, dt$
 

$\dfrac{dy}{dx} = \dfrac{(6t^2 - 6)\,dt}{(2t + 2)\,dt}$

$\dfrac{dy}{dx} = \dfrac{3t^2 - 3}{t + 1}$
 

When t = 0, dy/dx = -3
When t = 2, dy/dx = 3
When t = 5, dy/dx = 12

Answer: [ D ]