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Find all points (coordinates), if any, on the curve x = t³ + (5/2)t³ - 2t + 1 and y = 9t³ - 3t - 1 at which the tangent line is horizontal and vertical.

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Final answer:

To locate points with horizontal or vertical tangent lines on a parametric curve, we need to calculate derivatives of x and y with respect to t, and then find specific values of t where these derivatives are zero or undefined, respectively.

Step-by-step explanation:

The student is asking to find the points on the given parametric curve where the tangent line is horizontal or vertical. This involves calculating the derivatives of both x and y with respect to t and setting them equal to zero (for horizontal tangents) or checking where the derivative of y with respect to x is undefined (for vertical tangents).

First, we would find the derivative of x with respect to t (dx/dt) and the derivative of y with respect to t (dy/dt). Next, for horizontal tangents, we set dy/dt equal to zero and solve for t, then we use these values of t to find the corresponding x and y coordinates by plugging into the original equations for x(t) and y(t).

For vertical tangents, we find where dx/dt is equal to zero, solve for t, and again plug these values of t into the original equations to find the x and y coordinates. However, in the provided solution, this process is not applicable as it refers to points at specific times and positions, which seems to be taken from a different context, likely a physical scenario involving motion.

User Adam Badura
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