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A tangent line becomes horizontal when it lies completely parallel to the x-axis. This means that the slope of the line at that point is zero.
In terms of parametric equations, such as our example with

• \(x = t(t^2 - 3)\) and
• y \(y = 3(t^2 - 3)\),

we rely on derivatives to find where the tangent line becomes horizontal. Specifically, we look for where \( \frac{dy}{dt} = 0 \).
This derivative essentially tells us how the y-values are changing as t changes. Setting \( \frac{dy}{dt} = 0 \) implies no vertical change at that moment—hence a horizontal tangent.
In our example, \( \frac{dy}{dt} = 6t \). Setting this to zero gives us \( t=0 \), indicating the specific parameter value where the tangent to the curve is horizontal. Calculating back using the parametric equations, this results in a point on the curve, \((0, -9)\), where the tangent line is horizontal.

A vertical tangent line occurs when it is perfectly aligned with the y-axis, having an undefined or infinite slope. In parametric equations,

• x = \(t(t^2 - 3)\) and
• y = \(3(t^2 - 3)\),

this translates to observing the derivative \(\frac{dx}{dt}\).
If \(\frac{dx}{dt} = 0\), then the tangent line there could be vertical. This is because it indicates no change in x-values as t varies, causing the slope to approach infinity.
From our example parametric equations, we calculate \(\frac{dx}{dt} = 3t^2 - 3\). Solving \(3t^2 - 3 = 0\) yields \(t = 1\) and \(t = -1\), suggesting potential vertical tangent lines. Calculating each scenario into our parametric functions, we arrive at points \((-2, -6)\) and \((2, -6)\) respectively, where each line becomes vertical.

Parametric derivatives involve finding how variables change with respect to a shared parameter, usually t. When working with parametric equations like

• x = \(t(t^2 - 3)\) and
• y = \(3(t^2 - 3)\),

the process requires using the chain rule to compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
These derivatives help identify significant geometric features of curves defined parametrically. They allow us to determine characteristics such as slope by examining how x and y change in relation to t.
The derivative \( \frac{dy}{dx} \) is particularly important as it effectively captures the slope of the tangent line at any given point. Its evaluation is formulated as \( \frac{dy/dt}{dx/dt} \) when dealing with parametric forms.
For example, the derivatives yield \(\frac{dx}{dt} = 3t^2 - 3\) and \(\frac{dy}{dt} = 6t\) for our initial problem, providing insight into how both quantities of interest alter as t progresses. This critical information is necessary for recognizing horizontal or vertical tangents.