Find the equation of the tangent line of y=3x^2+1 through (0,1)

Question

Lost on this one.......

Answer 1

tangent line has slope = derivative = 6x

at 6x the point (0,1) has zero slope = 0 = 6(0)

the line goes through y=1 with slope =0

that means y=1 is the equation of the tangent line through the given point