You need to linearize f`(x) = 4x^2-5x+4 ` at `x = 6` using the following formula such that:

`y = f(6) + f'(6)(x - 6)`

You need to find f(6) substituting 6 for x in equation of the function f(x) such that:

`f(6) = 4*6^2 - 5*6 + 4 => f(6) = 118`

You need to evaluate f'(x) such that:

`f'(x) = (4x^2-5x+4)' => f'(x) = 8x - 5`

You need to evaluate f'(6) substituting 6 for x in equation of f'(x) such that:

`f'(6) = 8*6 - 5 = 43`

You need to substitute 118 for f(6) and 43 for f'(6) in equation of linearization such that:

`y = f(6) + f'(6)(x - 6) => y = 118 + 43(x - 6)`

**Hence, you may linearize the function `f(x) = 4x^2-5x+4` near `x = 6` using the equation `y = 118 + 43(x - 6).` **