Question

Find the points on the graph where the tangent is horizontal or vertical (dy/dx=(dy/dt)/(dx/dt))

x=t^3-3t^2, y=2t^2+8t-5

Answer 1

differentiate both functions with respect to t.

\[\frac{dx}{dt} = 3t^2 - 6t\]

\[\frac{dy}{dt} = 4t + 8\]

now using the hint

\[\frac{dy}{dx} = \frac{4t + 8}{3t^2 - 6t}\]

you may need to check if it can be simplified.

where is the gradient of the tangent horizontal...

when the 0 = 4t + 8

when is it vertical

when 3t^2 - 6t = 0

hope this helps

Answer 2

thank you for your clear reply! i followed your steps to get that the solutions for t are -2 (horizontal); 0,2 (vertical)

how do i check to make sure that the solutions are "real" or is this the end of the problem?

Answer 3

this is the end... the solutions to the vertical are asymptotes as you can't divide by zero...

Answer 4

ok thank you :)