Question

find the points on the curve where the tangent is horizontal or vertical. if you have a graphing device, graph the curve to check your work. (enter your answers as a comma - separated list of ordered pairs.)
$x = t^{3}-3t$, $y = t^{2}-5$
horizontal tangent $(x,y)=$
vertical tangent $(x,y)=$

Explanation:

Step1: Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)

Given \(x = t^{3}-3t\), then \(\frac{dx}{dt}=3t^{2}-3 = 3(t^{2}-1)=3(t - 1)(t + 1)\).
Given \(y=t^{2}-5\), then \(\frac{dy}{dt}=2t\).
The slope of the tangent line for parametric equations is \(\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2t}{3(t^{2}-1)}\).

Step2: Find horizontal - tangent points

For a horizontal tangent, \(\frac{dy}{dt}=0\) and \(\frac{dx}{dt}
eq0\).
Set \(\frac{dy}{dt}=2t = 0\), so \(t = 0\).
When \(t = 0\), \(x=t^{3}-3t=0\) and \(y=t^{2}-5=-5\).

Step3: Find vertical - tangent points

For a vertical tangent, \(\frac{dx}{dt}=0\) and \(\frac{dy}{dt}
eq0\).
Set \(\frac{dx}{dt}=3(t^{2}-1)=0\), then \(t=\pm1\).
When \(t = 1\), \(x=t^{3}-3t=1 - 3=-2\), \(y=t^{2}-5=1 - 5=-4\).
When \(t=-1\), \(x=t^{3}-3t=-1+3 = 2\), \(y=t^{2}-5=1 - 5=-4\).

Answer:

Horizontal tangent: \((0,-5)\); Vertical tangents: \((-2,-4),(2,-4)\)