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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
That's a good start. Now, find the midpoint of the line segment containing endpoints (0,0) and (630,0)
(315, 0) actually. Plot that point on your graph and draw a vertical line segment from the vertex to the midpoint of the base segment.
yea thats what i meant. and ok
That vertical line segment is essentially the axis of symmetry.
You should probably use the vertex form of a parabola to write the equation for it and make sure to include the correct restrictions for the domain.
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
So you should be able to express the equation of the parabola in vertex form. Is that all you're required to do for this problem?
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.