SOLUTION: A road is 32 feet wide is 0.4 foot higher in the center then it is on its sides. a) Find the equation of the parabola (Assume that the origin is at the center of the road)

Click here to see ALL problems on Quadratic-relations-and-conic-sections

Question 390339: A road is 32 feet wide is 0.4 foot higher in the center then it is on its sides.
a) Find the equation of the parabola (Assume that the origin is at the center of the road) - Show all work
b) How far from the center of the road is the road surface 0.1 foot lower than in the middle? -show all work.
please show work and thank you.
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
A road is 32 feet wide is 0.4 foot higher in the center then it is on its sides.
a) Find the equation of the parabola (Assume that the origin is at the center of the road)
Using ax^2 + bx + c = y
Since the center is at origin, (x=0) then c = +.4
x=-16; y=0
-16^2a - 16b + .4 = 0
256a - 16b = -.4
x=16; y=0
16^2a + 16b + .4 = 0
256a + 16b = -.4
:
add these two equations
256a - 16b = -.4
256a + 16b = -.4
-------------------adding eliminates b
512a = -.8
a =
a = -.0015625
:
Find b, it will equal 0, as you can see
-.0001562(256) + 16b = -.4
-.4 + 16b = -.4
16b = -.4 + .4
16b = 0
b = 0
Therefore:
The equation: -.0015625x^2 + .4 = y
:
:
b) How far from the center of the road is the road surface 0.1 foot lower than in the middle?
y = .3
-.0015625x^2 + .4 = .3
-.0015625x^2 = .3 - .4
-.0015625x^2 = -.1
x^2 =
x^2 = 64
x =
x = 8 ft from the center it is .1 ft lower than the middle

SOLUTION: A road is 32 feet wide is 0.4 foot higher in the center then it is on its sides. a) Find the equation of the parabola (Assume that the origin is at the center of the road) - Show