SOLUTION: Graph each parabola by hand. Give the vertex, axis, domain, and range. x = 2/3y^2 - 4y + 8 My work thus far: x = 2/3y^2 - 4y + 8 x - 8 = 2/3y^2 - 4y x - 8 = 2/3(

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Graph each parabola by hand. Give the vertex, axis, domain, and range. x = 2/3y^2 - 4y + 8 My work thus far: x = 2/3y^2 - 4y + 8 x - 8 = 2/3y^2 - 4y x - 8 = 2/3(      Log On


   

Question 841635: Graph each parabola by hand. Give the vertex, axis, domain, and range.
x = 2/3y^2 - 4y + 8
My work thus far:
x = 2/3y^2 - 4y + 8
x - 8 = 2/3y^2 - 4y
x - 8 = 2/3(y^2 - 6y)
x - 8 + 9 = 2/3(y^2 - 6y + 9) * After completing the square
x + 1 = 2/3(y - 3)^2
3/2(x + 1) = (y - 3)^2
Finally ends up in the right form, however, based on this equation my vertex would be: (-1,3), while the book answer is (2,3).. Any help would be appreciated!
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

x = 2/3y^2 - 4y + 8 

 x - 8 = 2/3y^2 - 4y 

 x - 8 = 2/3(y^2 - 6y)  Completing Square

 x-8 = %282%2F3%29%28y-3%29%5E2+-+highlight%28%282%2F3%29%2A9%29

x-8 = %282%2F3%29%28y-3%29%5E2+-+6

x-8 + 6 = %282%2F3%29%28y-3%29%5E2+

 x-2 = %282%2F3%29%28y-3%29%5E2+

x = %282%2F3%29%28y-3%29%5E2+%2B+2   V(2,3) axis is horizontal line y = 3

domain  x ≥ 2 Parabola opening right.  (Sketch it)

range: All real numbers



the vertex form of a Parabola opening right(a>0) or x=a%28y-k%29%5E2+%2Bh

 where(h,k) is the vertex and  y = k  is the Line of Symmetry