An equation of the tangent line to the curve y=2x^3-x^2+2 at the point (1,3) is?

So my approach is first calculate the derivative of y=2x^3-x^2+2 which is y'= 6x^2-2x, and then plug in (1,3) to get (1,4)

I know the correct answer using the y-yo=a(x-xo) and a should indicate the slope? But why is a 4 here? Can you explain that to me please?

Thank you very much for your help