Asked by Liz

Convert the curve to an equation in rectangular coordinates:
x=(t+2); y=-2sqrt(t)
Answered by Reiny
x = t+2 ----> t = x-2
y = -2√t
y^2 = 4t ---> t = y^2/4

then y^2/4 = x-2
y^2 = 4x - 8
Answered by Liz
On what intervals of t is the curve described by the given parametric equation concave up? Concave down?
x=t^2; y=t^(3) + 3t


I am a bit confused on how to solve this...any help/explanations are welcome!! (& greatly appreciated!)