Derivatives of e and ln x

Question

Let . Then the equation of the tangent line to the graph of  at the point  is given by  for

Answer 1

Is it y = f(x) = 4x +2 - 13e^x or y=f(x) = 4x+2-13ex ???

In the latter case, it is y = f(x) = (4-13e)x + 2

f(0) = 2, which does not agree with the data,

so my guess about the function is correct!!!

y' = 4 - 13e^x and so m=f'(0) = -9

B = y - mx = -11 - (-9)(0) = -11

y = -9x - 11