Chikae Y. answered • 01/13/23
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HS Math Teacher with 10+ years of teaching experience
Hi Rachel - your equation is correct. I'm not sure why you're being marked wrong... unless it's written as root(x3) or some other equivalent expression?
Chikae Y.
Good luck! I checked my work also on Desmos and on the Wolfram Alpha website and got the same answer still... lf there's supposed to be a different answer, please share :)
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01/13/23
Roger R.
tutor
x = (3t)^2 and y = (3t)^3.
The problem here is that the sign of y = (3t)^3 is negative for -∞ < t < 0, while y = x^(3/2) has only non-negative values. The correct equations are x^3 -y^2 = 0 or y^2= x^3 or x = ∛(y²), but not y = x^(3/2) !
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01/13/23
Chikae Y.
Ah I see - yes, I had assumed domain of t was t >=0, when I was checking my work - without an explicit restriction, I should have checked the negative values as well! Thanks for the note.
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01/13/23
Rachel M.
Thank you, I just wanted to make sure I wasn't going crazy before I emailed my professor.01/13/23