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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