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. $t = 0, -2$.

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!