# Pre - Cal

1. Find the vertex of the parabola: 4y^2+4y-16x=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.

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1. 4y^2+4y-16x=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)

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posted by Damon
2. (x+3)^2 = a(y-1)
(4+3)^2 = a (-5-1)

49 = -6 a
a = -49/6

x^2 + 6 x + 9 = -49/6 (y-1)

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

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posted by Damon
3. 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

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posted by Damon
4. ) 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

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