A building has an entry the shape of a parabolic arch 96 ft high and 18 ft wide at the base as shown below.
(picture will be in comments)
Find an equation for the parabola if the vertex is put at the origin of the coordinate system.

Question

A building has an entry the shape of a parabolic arch 96 ft high and 18 ft wide at the base as shown below. 
(picture will be in comments)
Find an equation for the parabola if the vertex is put at the origin of the coordinate system.

sleepyjess · Answer

http://prntscr.com/6bmg8e

e.mccormick · Answer

OK.  So, what formulas for Paraabolas do you know?

sleepyjess · Answer

This is what they gave for if the vertex is at the origin
x^2 = 4py

e.mccormick · Answer

OK.  Do you know what the p was?

e.mccormick · Answer

Oh, and on this one, you may need to find it from the three points they give you.

sleepyjess · Answer

3 points are 96, -9; -96, -9, 0, 0?

e.mccormick · Answer

Close, but not quite.  It never goes postitive in x.  (-96, -9), (-96, -9), and (0, 0).

e.mccormick · Answer

oops... I meant postitive in y....
wait, you also have the order wrong for (x,y)

e.mccormick · Answer

See if you can fix the points.  Remember, (x,y) form.

sleepyjess · Answer

96, -9
-96, -9
0, 0

sleepyjess · Answer

aahh
9, -96
-9, -96
0, 0

e.mccormick · Answer

Yah, (9, -96), (-9, -96), and (0, 0).  OK, so, can you make a calculation that hits those 3 points?  One nice things is the vertex is (0,0) so you know there are no shifts involved.

sleepyjess · Answer

I am completely stumped... :/

e.mccormick · Answer

OK.  Well, the basic parabola is $y=x^2$ right?

sleepyjess · Answer

yes

sleepyjess · Answer

They also give $x^2=-ay$. Would I need to use that?

e.mccormick · Answer

Well, if you think about it, the formula with all the options is:

$y=a(x+h)^2-k$ where a it the amplpitude and (h,k) is the vertex.

So, if you use what you know, you can fill that in.

sleepyjess · Answer

y = -96(x+0)^2 - 0?

e.mccormick · Answer

CLOSE!   but you do not know what a is.  You know what y is tho.

e.mccormick · Answer

And you know what the x linked to that y would be...

e.mccormick · Answer

Also, because (h,k) = (0,0) you can leave it off.

So for (x,y) = (9,-96) kist put it into

y=a(x)^2

Then solve for a.

sleepyjess · Answer

-96 = a*81
-1.89 = a

e.mccormick · Answer

yah.  So.  Now that you have an a, you can go back to:

$y=ax^2$ and replace a.

sleepyjess · Answer

y = -1.89(x+0)^2 - 0

e.mccormick · Answer

Yes, but the 0s do not matter and can go away.

And if it says it wants an exat answer, you may need to leave -96/81 in reduced fractional form.

sleepyjess · Answer

Sank chu :)

e.mccormick · Answer

So, get how that worked?

sleepyjess · Answer

Yes

e.mccormick · Answer

kk.  Have fun!

sleepyjess · Answer

haha

e.mccormick · Answer

With math, a lot of it really comes down to, "From this problem, what do I know and what formula could I put that into?"