Answer:
The equation of a line tangent to f(x) at = x - 3 is given by Y = 6x + 8
=> f'(x - 3) = 6x + 8
=> f(x - 3) = 3x^2 + 8x
=> f(x - 3) = 3(x^2 - 6x + 9) + 26(x - 3) + 51
=> f(x - 3) = 3(x - 3)^2 + 26(x - 3) + 51
=> f(x) = 3x^2 + 26x + 51
=> f'(x) = 6x + 26
Hope this helps!
:)
