SOLUTION: Eliminate the parameter. x = t2 + 3, y = t2 - 2 (2 points) y = x + 5, x ≥ 3 y = x2 - 5, x ≥3 y = x - 5, x ≥ 3 y = x2 + 5, x ≥3

Algebra ->  Trigonometry-basics -> SOLUTION: Eliminate the parameter. x = t2 + 3, y = t2 - 2 (2 points) y = x + 5, x ≥ 3 y = x2 - 5, x ≥3 y = x - 5, x ≥ 3 y = x2 + 5, x ≥3       Log On


   



Question 1137589: Eliminate the parameter.
x = t2 + 3, y = t2 - 2 (2 points)

y = x + 5, x ≥ 3

y = x2 - 5, x ≥3

y = x - 5, x ≥ 3

y = x2 + 5, x ≥3

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


NOTE: Use "^" (shift-6) to denote exponentiation....

Given: x = t^2 + 3, y = t^2 - 2

Rearranged: t^2 = x-3; t^2 = y+2 --> x-3 = y+2 --> y = x-5

Then, with the given definition x = t^2+3, we know the minimum value of x is 3; then the complete answer is

y = x-5; x >= 3