Expert: Richard J. Raridon - 3/14/2008

Question
I'm Emily and I'm currently 16 years old.
I have these 4 problems and I have absolutely no clue how to go about finding the vertex and which way they would open.

y²=128x
(x-5)² = -10(y+2)
y² - 4y - 8x + 25 = 0
4x - y² + 6y - 1 = 0

First the first problem, I figured I was supposed to square root it to get it into a y=ax form and since the square root of 128 is greater than 0, does that mean it opens up? That was as far as I could get :/

Answer
the y=ax+b form applies to a straight line, which these are not.  
the standard equation for a parabola opening right or left is      (y-k)^2 = 4a(x-h) where the vertex is (h,k).
so for y^2 = 128x, the vertex is at (0,0) and it opens to the right.  If you want to check it, put in a few values for x, calculate y, and plot x,y.  
the standard equation for a parabola opening up or down is
(x-h)^2 = 4a(y-k)
so for (x-5)^2 = -10(y+2) the vertex is is (5,-2) and it opens down.
re-write y^2-4y-8x+25 = 0 as (y-2)^2 = 8(x-21/8) so the vertex is at
(21/8,2) and it opens to the right
re-write 4x-y^2+6y-1 = 0 as (y-3)^2 = 4(x+2) so the vertex is at
(-2,3) and it opens to the right