The equation of a line tangent to f(x) at x-3 is given by Y=6x+8, what is the value of f'(x) ?

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!

:)

is should be y = 5x

step-by-step explanation:

correct me if im wrong