As the directrix is parallel to x-axis, the concavity of the parabola will be in vertical. Then, the general equation is in the form: \((x-x_0)^2=2p(y-y_0)\), where \((x_0,y_0)\) are the coordinates of the vertice and p is the parameter (distance between the focus and the directrix). The parameter is \(p=1\). The vertice is the midpoint between the focus and the projection of the focus in the directrix. Then, the vertice is \((6,2-\dfrac{1}{2})=(6,\dfrac{3}{2})\). Hence, the parabola is: \((x-6)^2=2\cdot1\cdot(y-\dfrac{3}{2})\\\\x^2-12x+36=2y-3\\\\2y=x^2-12x+39\\\\\boxed{y=\frac{1}{2}x^2-6x+\dfrac{39}{2}}\)
Now, derivating:
\(y'=\frac{1}{2}\cdot2x-6\\\\\boxed{y'=x-6}\)
