Find the equation of the tangent line (*in xy-coordniates) to the curve given parametrically by x=t^3-t and y=t^2-1 when t=-1
Added by Victoria N.
Step 1
Given t = -1, we can find the corresponding x and y values by plugging t into the parametric equations: x = (-1)^3 - (-1) = -1 + 1 = 0 y = (-1)^2 - 1 = 1 - 1 = 0 So, the point on the curve is (0, 0). Show more…
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