Pre  Cal
posted by Veronica .
1. Find the vertex of the parabola: 4y^2+4y16x=0
2. Find an equation of the parabola, opening down, with vertex (3,1) and solution point (4,5).
3. The main cables of a suspension bridge are 20 meters above the road at the towers and 4 meters above the road at the center. The road is 80 meters long. Vertical cables are spaced every 10 meters. The main cables hang in the shape of a parabola. Find the equation of the parabola. Then, determine how high the main cable is 20 meters from the center.
Please help, I've tried solving the problems many times and still can't get it right.

4y^2+4y16x=0
y^2 + y = 4 x
y^2 + y + 1/4 = 4 (x + 1/16)
(y+1/2)^2 = 4 (x+16)
vertex at (16 , 1/2) 
(x+3)^2 = a(y1)
(4+3)^2 = a (51)
49 = 6 a
a = 49/6
x^2 + 6 x + 9 = 49/6 (y1)
x^2 + 6 x + 54/6 =  49 y/6 + 49/6
x^2 + 6 x  5/6 = 49 y/6
6 x^2 + 36  5 = 49 y 
start at the middle up 4 end up 20 (16 above middle
y = kx^2
16 = k(40)^2
16 = k (4)^2(10^2)
1 = 100 k
k = .01
so , adding the 4 at the middle
y = 4 + .01 x^2
at x = 20
y = 4 + .01 (4)(100) = 8 
) The main cables of a suspension bridge are 20 meters above the road at the towers and 4 meters above the road at the center. The road is 80 meters long. Vertical cables are spaced every 10 meters. The main cables hang in the shape of a parabola. Find the equation of the parabola. Then, determine how high th