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find an equation of the tangent line of the graph of y=(x2+3)/(x-1), x=2

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Given curve is y \(=\frac{x^2+3}{x-1}\)

Slope of tangent to given curve is m = dy/dx \(=\frac{(x-1)\times2x-(x^2+3)\times1}{(x-1)^2}\)

Slope of tangent to curve at x = 2 is m \(=\frac{1\times4-7}{1^2}=\frac{4-7}{1}\) = -3

Ordinate of point P on curve when x = 2 is y \(=\frac{2^2+3}{2-1}\) = 7/1 = 7

Hence, point (2, 7) lies on curve.

Thus tangent of given curve has slope  -3 and passing through point (2, 7) when x = 2.

∴ Equation of tangent at x = 2 is y - y1 = m(x - x1)

⇒ y - 7 = -3(x - 2)  (∵ y1 = 7, x1 = 2, m = -3)

⇒ y - 7 = -3x+6

⇒ 3x+y-7-6 = 0

⇒ 3x+y = 13

Hence, equation of required tangent is 3x+y = 13.

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