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the height of the arch is 630ft as well as the width

this was the assignment. https://i.gyazo.com/ea2c7cf9f897771440fb99ec9074fb06.png

Start by making a sketch of it on graph paper.

do i just place the vertex at the origin?

It's probably best to place one of the ends of the foundation at the origin.

ok give me a second to sketch it

|dw:1439572894166:dw|

???

NO ONE?

(0, 315)

yea thats what i meant. and ok

That vertical line segment is essentially the axis of symmetry.

|dw:1439574756537:dw|

|dw:1439574890871:dw|

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

Yes that. You already know the values for (h,k).

what do i plug in for a though? do i leave it as it is?

You have to solve for a

To do that, insert (h,k), then for (x,y) insert (0,0)

Then solve for a

oh ok ill do that

i got a= -2/315

Show the work you did to arrive at that result

a(0-315)^2 + 630 =0
99225a + 630 = 0
99225a = -630
a = -630/99225
a = -2/315

Okay good.

Actually, that value for a is an approximation. It's not exact.

yea thats pretty much it. My teacher just told us to do that for one of the pictures given.

thanks for the help.

yw