Need some help with how I am doing this problem.

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. 
x = 2t³ + 3t² − 12t, y = 2t³ + 3t² + 3

 

I think what I have been doing wrong is mixing up which is the horizontal and vertical. If anyone could help me clarify the steps that would be great! Thank you!

Answer 1

This is a parametric equation, where t is the parameter. The function of interest is y= f(x) and the derivative of y with respect to x can be derived from:

 

dy/dt= (dy/dx)(dx/dt)

 

or

 

dy/dx= (dy/dt)/(dx/dt)= (6t^2 + 6t)/(6t^2 + 6t -12)

 

Since this derivative is zero when t=0, it appears there is a horizontal tangent at the point:

 

(x,y)= (0,3)

 

If you factor the denominator, you will find it is equal to 6(t+2)(t-1), which means there will be a vertical tangent at t= 1. Where t=1. You can plug in t=1 into the x and y parametric equations to find the tangent points for x and y.

 

You might want to verify this with a graphing calculator as suggested.

 

- Ben