find the points on the curve where the tangent is horizontal or vertical. if you have a graphing device…

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.)\n$x = t^{3}-3t$, $y = t^{2}-5$\nhorizontal tangent $(x,y)=$\nvertical tangent $(x,y)=$

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.)\n$x = t^{3}-3t$, $y = t^{2}-5$\nhorizontal tangent $(x,y)=$\nvertical tangent $(x,y)=$

Answer

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}\neq0). 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}\neq0). 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))