Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.x = t4 + 5, y = t3 + t; t = −1
Added by Emilia R.
Step 1
Given x = t^4 + 5 and y = t^3 + t, when t = -1: x = (-1)^4 + 5 = 1 + 5 = 6 y = (-1)^3 + (-1) = -1 - 1 = -2 Therefore, the point on the curve is (6, -2). Show more…
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