Given that the graph of f(x) passes through the point (6,7) and that the slope of its tangent line at (x,f(x)) is 3x+2, what is f(2)?
so dy/dx = 3x + 2
y = (3/2)x^2 + 2x + c
but (6,7) lies on it, so
7 = (3/2)(36) + 2(6) + c
7 = 54 + 12 + c
c= -59
y = f(x) = (3/2)x^2 + 2x - 59
f(2) = (3/2)(4) + 4 - 59 = -49
check my arithmetic