Question

Can someone please explain this to me?
A tunnel is in the shape of a parabola. The maximum height is 31 m and it is 13 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 31 meters and its width from left to right is 13 meters.
(I'll post the picture!)
What is the vertical clearance 2 m from the edge of the tunnel?

Answer 1

find equation of parabola, then plug in x = 4.5

Answer 2

Oh! Okay, hang on.

Answer 3

Would it be..\[\frac{ x^2}{ 42.25 }+\frac{ y^2 }{ 961 }=1\] ?

Answer 4

whoa hold on.... that is an ellipse not a parabola, the "y" should not be squared

Answer 5

-_- As you can see, I suck at these.. So the x and y aren't squared, and should it be subtraction instead of addition, too?

Answer 6

we know the x-intercepts are -6.5 and 6.5

\[y = a(x+6.5)(x-6.5)\]

where a is slope of parabola and can be found using max height at point (0,31)
\[31 = a (6.5)(-6.5)\]

Answer 7

Oh wait, ellipses are the ones that have (x-h)^2, and (y-k)^2, aren't they?

Answer 8

yeah this is a parabola not an ellipse :)

Answer 9

Wait, what?

Answer 10

I'm confused. If this is a parabola, then aren't the x and y supposed to be squared? You lost me.

Answer 11

@dumbcow @sourwing

Answer 12

It is a parabola

Answer 13

I feel retarded. This is my very last test in this Pre-Cal Honors class, and I need a 100, because I only have a 91...

Answer 14

So how do I make the equation. I'll remember if someone will just refresh my memory.

Answer 15

parabola

https://www.khanacademy.org/math/algebra/quadratics/solving_graphing_quadratics/v/graphing-a-parabola-using-roots-and-vertex

Answer 16

\[y = ax^2 +bx +c\]
if x-intercepts are known, write in factored form
\[y = a(x-p)(x-q)\]

Answer 17

What are p and q?

Answer 18

the x-intercepts ... (p,0) and (q,0)

Answer 19

So -6.5 and 6.5. Then what the heck are the x's?

Answer 20

And what's a?

Answer 21

l already explained "a" is slope of parabola, it determines the shape, how wide or narrow the graph is

Answer 22

I found something similar on my school page, and I got a=-1.36.

Answer 23

\[y = -\frac{31}{6.5^2}(x-6.5)(x+6.5)\]
plug in x = 4.5 and that gives the height 2 m from edge which is the "y" value on the parabola

Answer 24

no to find "a:, plug in point for max height (0,31)

\[31 = a(0-6.5)(0+6.5)\]

Answer 25

Ugh, FLVS did it the opposite way. So it's .73?

Answer 26

yeah sorry im not familiar with FLVS methods

Answer 27

I took a screenshot, but I don't know how to save it. Hang on.