Question #34c6f

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Answer 1 · 2016-07-19T10:54:51

horizontal tangents #t = pm 2# .

vertical tangents #t = 1#

well #dy/dx = (dy/dt)/(dx/dt)#

# = (3t^2 - 12)/(2t-2)#

you will have horizontal tangents when the numerator = 0

ie #3t^2 = 12, t = pm 2# ..... and just check that he denominator is not also 0 at #t = pm 2#, as that might cause problems

you will have vertical tangents when the denominator = 0

ie #2t=2, t = 1#