ln(x^2 - 3x + 1) = y at the given point (3,0)
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Derivative of ln(x^2 - 3x + 1) = 1/(x^2 - 3x + 1) * (2x - 3)
Plug in 3 to find the slope of the tangent line at (3, 0):
1/(3^2 - 3(3) + 1) * (2*3 - 3) = 3
y - 0 = 3(x - 3)
y = 3x - 9 <---answer
Plug in 3 to find the slope of the tangent line at (3, 0):
1/(3^2 - 3(3) + 1) * (2*3 - 3) = 3
y - 0 = 3(x - 3)
y = 3x - 9 <---answer
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f ` (x) = (2x - 3) / (x^2 - 3x + 1)
f ` (3) = 3
y = 3 ( x - 3 )
y = 3 x - 9
f ` (3) = 3
y = 3 ( x - 3 )
y = 3 x - 9