Originally Posted by
scottydoint Sorry for all of these questions, but this is last minute and my teacher can't even get these problems right in class so I am trying to teach myself from a book that is hard to follow.
How do you graph this parabola:
x=-3y^2-6y+3
Mr F says: The standard form is found by completing the square:
$\displaystyle {\color{red}x = -3 (y^2 + 2y - 1) = -3 [(y+1)^2 - 2] = -3(y+1)^2 + 6} \, ....$
and how do you graph this circle? I don't think its possible.
x^2+(y-1)^2+1=0 Mr F says: $\displaystyle {\color{red} \Rightarrow x^2 + (y-1)^2 = -1}$ which is clearly not possible for real values of x and y.