The discussion revolves around the behavior of a graph defined by parametric equations, specifically addressing the occurrence of both vertical and horizontal tangents at the point (2, -4). Participants explore the implications of the tangents in relation to the parameter t, examining how the graph crosses itself at this point.
Participants generally agree on the existence of both vertical and horizontal tangents at the point (2, -4) due to the graph crossing itself, but there is some uncertainty regarding the conditions under which these tangents occur and the calculations involved.
Some participants express confusion about the computation of tangents, and there is a lack of clarity on the mathematical steps leading to the identification of tangents at the specified point.
Just show us your work, not an image of it, especially one that doesn't render.Calpalned said:https://mail.google.com/mail/u/0/?ui=2&ik=263fc3780d&view=fimg&th=14b0f97b7ad0fc5a&attid=0.1&disp=inline&safe=1&attbid=ANGjdJ9rFhryHaxukeJKxVFkvC2oxwR748Jb1IOpuC3WKs59gfmKcuR1B_n8TTjb5NhFNTaqw8vUgJYVuaXRo63FYUiD7TCzRXcrnGNGm3MDzoHF8zC1VlC2vVLwC3o&ats=1421902241584&rm=14b0f97b7ad0fc5a&zw&sz=w1273-h532
When t = 2, x = 2 and y = -4, just as you say. And dy/dx = 0 when t = 2, so the tangent is horizontal at (2, -4). Why do you think that the tangent is vertical when t = 2?Calpalned said:For the parametric equations x = t^3 - 3t and y = t^3 - 3t^2 I got that the graph has a vertical tangent when t is = to positive or negative one. And it is horizontal at t = 2. However, this implies that at the point (x,y) = (2, -4) the graph has both a vertical and horizontal tangent. How is this possible? Thanks
Right, but they occur for two different values of t.Calpalned said:My friend from math class actually explained it to me, but thank you for your help. At the point (2, -4), the graph crosses itself, so as a result there are two tangents at that spot.