x = t^2 - t^-1

y = 1 +t ^2

at the point corresponding to t = 1

Please show work

The slope of the tangent line is y'(t)/x'(t). Here this equals to

m = 2t/(2t + t^(-2)). For t = 1 m = 2/(2+1) = 2/3.

The tangent line goes through the point (x(1); y(1)) = (0; 2).

So the equation is

y - 2 = (2/3)x or y = (2/3)x + 2.

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