For what values of t does the curve given by the parametric equations x = t^3-t^2-1 and y = t^4+2t^2-8t have a vertical tangent

Hello fellow user. Remember that vertical tangents always occur wherever the derivative of your curve is undefined, likewise, horizontal tangents always occur wherever your curve is zero. Now, the way you set up these kind of problems is your realize, what is it that I'm looking for? Now, what your looking for is wherever the slope of your curve is undefined, so you want to know where your dy/dx is undefined. Now, knowing that we need to know that dy/dx is undefined is great and all, but we now need to be able to work with this in a way such that it is defined explicitly in terms of the parameter t, rather than implicitly, where it is simply known that we possess the derivative of some function y(t) with respect to x(t). Anyways, the way you set this up is in the following way, dy/dx is mathematically equivalent to (dy/dt)/(dx/dt), and although this formulation seems simple, it allows us to be able to focus on the derivatives of parts explicitly defined in terms of our parameter t. Now, what we really care about however in this case, is simply where the denominator is equal to 0, because that is where our derivative will be undefined. This means all we need to do is differentiate the x term, set it equal to 0 and solve. Once you do that you have the equation 3t^2-2t=0. Factoring you get, t(3t-2)=0 where the solutions come out to t=0 and t=2/3. 
I hope that this answer helps and that you have a great day
Kirkland Tucker

The expression of this would be 3.75x+y= 23