`y = ln(x^2 - 3x + 1), (3,0)` Find an equation of the tangent line to the curve at the given point.

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The equation of the tangent line to the curve `y = ln (x^2 - 3x + 1)` , at the point (3,0) is the following, such that:

`f(x) - f(x_0) = f'(x_0)(x - x_0)`

You need to put ` ` `f(x) = y, f(x_0) = 0, x_0 = 3`

You need to evaluate f'(3),...

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