You need to remember what is the form of equation of the tangent line to the graph of the function, at the point x = 3, such that:

`y - f(3) = f'(3)(x - 3)`

You need to determine f(3) substituting 3 for x in equation of teh function such...

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You need to remember what is the form of equation of the tangent line to the graph of the function, at the point x = 3, such that:

`y - f(3) = f'(3)(x - 3)`

You need to determine f(3) substituting 3 for x in equation of teh function such that:

`f(3) = 9e^(27)-12`

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

`f'(x) = 9e^(3x^2)*(3x^2)' - 4 => f'(x) = 54x*e^(3x^2)`

You need to determine f'(3) substituting 3 for x in equation of teh function such that:

`f'(3) = 54x*e^(3x^2) => f'(3) = 162*e^27`

You may write the equation of tangent line such that:

`y -9e^(27)+12 = 162*e^27(x - 3)`

You need to put the equation in slope intercept form, such that:

y = 9e^27 -12 + 162x*e^27 - 486*e^27

`y = 162x*e^27 - 477*e^27 - 12`

**Hence, evaluating the equation of tangent line yields `y = 162x*e^27 - 477*e^27 - 12` where `m = 162*e^27 ` and `b =- 477*e^27 - 12.` **