College algebra
posted by Jamie on .
Find the maximum value of: y=3x^2 15x +1

in any parabola ax^2+bx+c=0, the max value is attained at x = /2a.
So, here that would be when x = 15/6 = 5/2
y(5/2) = 19.75 
another way of solcinf the problem is by deriving and setting dy/dx=0...but the above sol. is easier that but its disadv. is you may forget the formula.. :D