The equation below represents a curve, c

y=3x^2-18x+5

a. find dy/dx and d^2y/dx^2

b. find the coordinates of the stationary points on the curve

c. determine the nature of the stationary points.

provide full working out pls.

y=3x^2-18x+5
a. find dy/dx and d^2y/dx^2
dy/dx = 3*2x - 18 = 6x - 18
d^2y/dx^2 = 6

b.find the coordinates of the stationary points on the curve 
stationary points of a curve are those where its derivative is zero:
dy/dx = 6x - 18 = 0
6x = 18
x = 3
and substituting in the original equation:
y = 3x^2 - 18x + 5
y = 3(3)^2 - 18(3) + 5
y = 3*9 - 54 + 5
y = 27 - 49
y = -22
therefore the stationary point is (3, -22)

c. determine the nature of the stationary points.
the second derivative at (3, -22) equals 6, which is greater than zero, therefore by the second derivative test, the point is a local minimum