Question

How to put this parabola in standard equation form?

Answer 1

I have to use the measurements of the Gateway Arch which are that it is 630 feet tall and 630 wide at the base.

Answer 2

Translate this word problem into numerical data: The vertex of this parabola is (0,630). When x=315, y=0: (315,0) is a point on the parabola.

Standard form would be either y=ax^2 + bx + c, or the "vertex form," y=a(x-h)^2 + k, where (h,k) is the vertex.

Answer 3

Could you explain how you got 315?

Answer 4

And I believe it wants the vertex form.

Answer 5

The parabola opens down, right? In doing so, it intersects the x-axis in two different places. Since the parabola is symmetric with respect to the y-axis, and since the two "feet" of the parabola are 630 feet apart, each foot intersects the x-axis (i. e., the ground) 315 feet from the y-axis. Thus, the 2 x-intercepts are (-315,0) and (315,0).

Answer 6

Identify the coordinates of the vertex: (h,k)

Answer 7

Oh that makes sense. h would be 0 and k would be 630?

Answer 8

Yes, that's right.

Answer 9

So, which letters in the vertex formula could be replaced by knowns?

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

Answer 10

y=a(x)^2 + 630?

Answer 11

If you do this correctly, you'll have only one unknown left to find / calculate. Which parameter is that, and how'd you find it?

Answer 12

I need to calculate a but I'm not sure what the represents.

Answer 13

that*

Answer 14

y=a(x)^2 + 630 is correct, but you have 3 unknowns. Find the constant, ' a . '

Answer 15

You have 2 points on the parabola, don't you? One is the vertex, (0,630); the other is the location of one of the "feet" of the parabolic arch: (315,0). This info is sufficient to enable you to find a.

Answer 16

I understand the two points we have but I'm not sure how to apply that in finding a.

Answer 17

The equation we're working with is \[y=a(x-h)^2+k\]

Please make four substitutions, as follows: h=0, k=630, y=0, x=315. Find a.

Answer 18

y = a(315)^2 + 630
y = a(99225) + 630

Answer 19

0 = a(99225) + 630
-630 = a(99225)
a = -.006349

Answer 20

What happened to y=0? One of the "feet" of the parabola is (315,0).

Answer 21

all right. Now, check your work. substitute your 'a' and your (0,630) into \[y=a(x-h)^2+k.\]

Answer 22

Then let x=315. Does your equation correctly predict that y will be 0?

Answer 23

0 = -.006349(99225) + 630
0=0

Answer 24

Yes it does I think

Answer 25

So, what would you conclude at this point?

Answer 26

The equation is correct?

Answer 27

Yes. I used your value for 'a' and obtained a result of 0.02, which was likely due to round-off error in your value of a.

But it seems to me that you have learned the gist of this problem and are OK.
Any questions?

Answer 28

I think I understand it a lot better now. Thank you for all your help!

Answer 29

My pleasure, Alex! Take care. Over and out.