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.

- anonymous

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- anonymous

@phi

- anonymous

@UsukiDoll

- phi

do you know the equation of a parabola in "vertex form" ?

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## More answers

- anonymous

is it (x-h)^2? i kind of forget

- anonymous

hello?

- phi

look in your notes

- anonymous

ok hold on

- anonymous

y = a(x-h)^2 + k

- anonymous

My answer choice are:
18 m
32 m
46 m
4 m
I was thinking I could approximate at 32m...

- 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

actually the turning point, or vertex is at the origin

- 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

(0,50)

- phi

if you put those numbers into the equation what do we get ?

- anonymous

y = a(x-h)^2 + k
y = a(x-0)^2 + 50

- anonymous

for my problem the equation would simplify to
y = a(x)^2 right?

- 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

**y = a(x)^2 right?***
you mean
y= a x^2 + 50

- anonymous

i was saying for my equation...

- anonymous

with the parable i am supposed to be working with. just making sure i am understanding.

- anonymous

also trying to show that i am understanding what you are saying

- phi

let's use my picture, then we will go back and use your picture.

- anonymous

ok got it

- anonymous

y = ax^2 + 50
0 = a(5)^2 + 50

- anonymous

0 = 25a + 50
-50 = 25a
-2 = a

- 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

3

- phi

now we use x=3 in the equation and find y. y will be the vertical height

- anonymous

y= -2 x^2 + 50
y = -2 (3)^2 + 50
y = -2 (9) + 50
y = -18 + 50
y = 32

- 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

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

haha ok thanks

- anonymous

please give me a medal @saseal

- 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

ok

- 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

|dw:1439210841995:dw|

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