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To determine where the curve described by the parametric equations:

\[
x = 3t^3 - t^2 + 2t, \quad y = t^3 + 3t^2 - 4
\]

has a horizontal tangent, we need to follow these steps:

### Step 1: Horizontal Tangent Condition
A curve has a horizontal tangent when the derivative of \( y \) with respect to \( t \), \( \frac{dy}{dt} \), is zero.

### Step 2: Find \( \frac{dy}{dt} \)
The \( y \)-equation is:

\[
y = t^3 + 3t^2 - 4
\]

Differentiate \( y \) with respect to \( t \):

\[
\frac{dy}{dt} = 3t^2 + 6t
\]

Set \( \frac{dy}{dt} = 0 \) to find the values of \( t \) where the tangent is horizontal:

\[
3t^2 + 6t = 0
\]

### Step 3: Solve for \( t \)
Factor the equation:

\[
3t(t + 2) = 0
\]

This gives:

\[
t = 0 \quad \text{or} \quad t = -2
\]

### Step 4: Verify Answer
Now that we've found \( t = 0 \) and \( t = -2 \), we check which answer choice corresponds to these values.

Answer choice **1** \( t = 0, -2 \) is correct.

### Conclusion:
The correct answer is **1. \( t = 0, -2 \)**.

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Would you like a more detailed explanation on how derivatives relate to tangent lines, or have any other questions?

Here are five related questions to explore further:
1. How do you find vertical tangents for parametric curves?
2. Can a point on a parametric curve have both a vertical and horizontal tangent?
3. What is the geometric interpretation of a parametric curve?
4. How do you find the second derivative for parametric curves?
5. What happens when both \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) are zero?

**Tip**: When solving for tangents in parametric equations, always check both \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) for vertical and horizontal tangents!

Determine all values of t for which the curve given parametrically by x = 3t^3 - t^2 + 2t, y = t^3 + 3t^2 - 4 has a horizontal tangent?

Determine where the parametric curve defined by x = 3t^3 - t^2 + 2t and y = t^3 + 3t^2 - 4 has horizontal tangents. This solution involves differentiation and solving dy/dt = 0. Suitable for high school or early college students.

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