anonymous
  • anonymous
MEDAL AND FAN!!! A tunnel is in the shape of a parabola. The maximum height is 50 m and it is 10 m wide at the base as shown below. A parabola opening down with vertex at the origin is graphed on the coordinate plane. The height of the parabola from top to bottom is 50 meters and its width from left to right is 10 meters. What is the vertical clearance 2 m from the edge of the tunnel? I will post a picture in a second.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@phi
anonymous
  • anonymous
@UsukiDoll
phi
  • phi
do you know the equation of a parabola in "vertex form" ?

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anonymous
  • anonymous
is it (x-h)^2? i kind of forget
anonymous
  • anonymous
hello?
phi
  • phi
look in your notes
anonymous
  • anonymous
ok hold on
anonymous
  • anonymous
y = a(x-h)^2 + k
anonymous
  • anonymous
My answer choice are: 18 m 32 m 46 m 4 m I was thinking I could approximate at 32m...
phi
  • phi
where (h,k) is the vertex (in this case the "top" of the parabola) next do you have a picture ? with any luck you could sketch it. it would look like this |dw:1439209730473:dw|
anonymous
  • anonymous
actually the turning point, or vertex is at the origin
phi
  • phi
the reason to study math is to practice "thinking". (or "puzzle things out" if you like) we can draw the picture any way we like. if we use my picture, what are the coordinates of the vertex?
anonymous
  • anonymous
(0,50)
phi
  • phi
if you put those numbers into the equation what do we get ?
anonymous
  • anonymous
y = a(x-h)^2 + k y = a(x-0)^2 + 50
anonymous
  • anonymous
for my problem the equation would simplify to y = a(x)^2 right?
phi
  • phi
ok and x-0 can be simplified to x, so y= ax^2 + 50 next, notice we are told the width (at ground level ) is 10 and with the origin at the middle, one side is at (5,0) put those numbers in for x and y and "solve for a"
phi
  • phi
**y = a(x)^2 right?*** you mean y= a x^2 + 50
anonymous
  • anonymous
i was saying for my equation...
anonymous
  • anonymous
with the parable i am supposed to be working with. just making sure i am understanding.
anonymous
  • anonymous
also trying to show that i am understanding what you are saying
phi
  • phi
let's use my picture, then we will go back and use your picture.
anonymous
  • anonymous
ok got it
anonymous
  • anonymous
y = ax^2 + 50 0 = a(5)^2 + 50
anonymous
  • anonymous
0 = 25a + 50 -50 = 25a -2 = a
phi
  • phi
yes, so the equation is y= -2 x^2 + 50 next we have to interpret **vertical clearance 2 m from the edge *** all the way to the edge is 5 meters. if we move 2 m to the left, what x value is that ?
anonymous
  • anonymous
3
phi
  • phi
now we use x=3 in the equation and find y. y will be the vertical height
anonymous
  • anonymous
y= -2 x^2 + 50 y = -2 (3)^2 + 50 y = -2 (9) + 50 y = -18 + 50 y = 32
anonymous
  • anonymous
oh lol so if this was the parabola in the equation the answer would be 32? I think this is also the same for my parabola because we have the same dimensions. I will check
phi
  • phi
looks like your guess way up top was a good one. if we draw the parabola with the vertex at (0,0), it looks like this |dw:1439210510774:dw|
anonymous
  • anonymous
haha ok thanks
anonymous
  • anonymous
please give me a medal @saseal
phi
  • phi
yes, we will get the same answer. with the origin at the vertex, the equation is y= a x^2 now we put in the point (5,-50) and solve for a -50= a*5*5 a= -2 y= -2 x^2
anonymous
  • anonymous
ok
phi
  • phi
next, we put in x=3 to find y y= -2 *3*3 y= -2*9= -18 now we have to be careful. that is not the height. that is the distance down from the vertex the height above the "bottom" at y= -50 is the difference between -18 and -50
phi
  • phi
|dw:1439210841995:dw|

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