Application of Differentiation

1a. y = (2x+1)(x-3) = 2x^2 - 5x - 3
dy/dx = 4x - 5

When the parabola cuts the y-axis, x=0,
thus the slope of tangent at (x=0) = 4(0)-5 = -5
the slope of normal at (x=0) = -1/-5 = 1/5

At x=0, y=-3, thus the point of intersection is (0, -3).

Therefore the equation of normal at (0, -3) is
y+3 = 1/5 x, i.e. x-5y-15=0.

b. When the point has zero gradient, dy/dx=0, x=5/4
and also y=2(5/4)^2 - 5(5/4) - 3 = -49/8.
So the point of zero gradient is (5/4, -49/8).

At that point, the equation of tangent is y=-49/8 (horizontal),
and the equation of normal is x=5/4 (verticle).