For my chapter 12 parabola project, I will be studying the St.Louis Gateway Arch, in St.Louis, Missouri.

The St.Louis Arch was the brain-child of Eero Saarinen and Fred Severud. It was built between February 12, 1963 and October 28, 1965. It is 630 feet tall and spans 630 feet from width of the base. The arch has 60 foot deep foundations to withstand a max of 150 mile per hour winds, making the structure sway 18 inches. The cost of the arch was estimated to be $13 million consisting of 900 tons of stainless steel, weighing a total of 17,246 tons!

"The St. Louis Gateway Arch is in the form of an inverted catenary, which is a very stable structure that is often used in bridges, domes, and arches.
A catenary is the shape that a chain or necklace forms when held by the two ends.The Dutch mathematician Christiaan Huygens named this curve from the Latin word catenarius, which means "related to a chain."The equation for a catenary curve is: y = k cosh(x/k), where cosh is the hyperbolic cosine [cosh(x) = (ex + e-x)/2] and k is the y-intercept (where the curve hits the y-axis)." credits to http://www.enchantedlearning.com/history/us/monuments/stlouisarch/

The parablola of the St.Louis Arch opens down, and has a vertex of (315, 630), which is a maximum. The vertex in this situation is the maximum hight the St.Louis Arch reaches. I found the vertex by graphing the parabolic shape of the arch. Since the arch is spans 630 feet as width, the x-axis has to span at least to 630 or more. Since the max height of the parabola is also 630, the y-axis would have to be at least 630 or more tall. The St.Louis Arch does have 60 foot deep foundations keeping the structure stable, so as a negative factor, the parabola will barely cross over the x-axis, which represents the gound, since the foundations are burried underground. The equation for the axis of symmetry for the St.Louis Arch is X=315

Equations:

Vertex Form: y-630= -0.0063(x-315)
Factored Form: y= -0.0063(x-0)(x-630)
Standard Form: y= -0.0063x^2+4x+0
Using the program Quadratic Regression Applet, I was able to figure out the standard for equation for the St.Louis Arch on a graph.
I plugged in all of the numbers and figured out the other equations by figuring out what the other variables stood for.

The y-intercept of my graph would be (0,-60) but the graph from EXCEL is derpy... The y-intercept is (0,-60) because i set the graph so the on leg of the arch would represent the y-axis. It is negative because as stated before, the arch has 60 foot foundations undergound, when the ground is considered the x-axis. I basically found it by graphing the parabola.

Discriminant:

16-4(-0.0063)x0= 16
Since the answer is positive,it means the parabola crosses the x-axis twice, that it is factorable, and since 16 is a perfect square, the two solutions are rational numbers.

X-Intercepts:

My parabola has two x-intercepts because the x-axis would represent gound level in my graph, and the arch has 60 foot deep foundations underground. They are (0,0) and (0,630) I found them by graphing the standard form equation on abettercalculator.com but you can find them without a graph by using the vertex form
(y-630= -0.0063(x-315) ) for the vertex, the factored form (y= -0.0063(x-0)(x-630) ) for one x-intercept, and the standard form (y= -0.0063x^2+4x+0) to find the other x-intercept. The x-intercepts represent when the legs of the arch hit ground level. Factoring the parabola would give you two binomials that would simplify into the standard form equation of the parabola which then can be revealed to the other forms of equations which will eventually give you the x-intercepts.

QUESTION!!!

Inside the arch are a series of spherical elevators that take you from 60 feet below the arch, to nearly the top of the arch where the observatory is. We want to find out how long it would take (in seconds) for the elevator to be 420 feet from the starting point. When will the elevator be 420 feet from the starting point? (hint, two solutions)

Since we are looking for when the elevator is 420 feet from the starting point, y=420
Fit in with the standard for equation for the arch, the equation is 420= -0.0063x^2+4x+0
In order to put this into a quadratic equation, we need 420 to be on the other side, so subtract. y=-0.0063^2+4x-420
Now we have our a, b and c numbers, we can plug it all into the quadratic equation.

16 +- sqrt(-4^2 - 4(-0.0063)(-420)) all over 2(-0.0063)

## St.Louis Gateway Arch

For my chapter 12 parabola project, I will be studying the St.Louis Gateway Arch, in St.Louis, Missouri.

The St.Louis Arch was the brain-child of Eero Saarinen and Fred Severud. It was built between February 12, 1963 and October 28, 1965. It is 630 feet tall and spans 630 feet from width of the base. The arch has 60 foot deep foundations to withstand a max of 150 mile per hour winds, making the structure sway 18 inches. The cost of the arch was estimated to be $13 million consisting of 900 tons of stainless steel, weighing a total of 17,246 tons!

"The St. Louis Gateway Arch is in the form of an inverted catenary, which is a very stable structure that is often used in bridges, domes, and arches.

A catenary is the shape that a chain or necklace forms when held by the two ends.The Dutch mathematician Christiaan Huygens named this curve from the Latin word catenarius, which means "related to a chain."The equation for a catenary curve is: y = k cosh(x/k), where cosh is the hyperbolic cosine [cosh(x) = (ex + e-x)/2] and k is the y-intercept (where the curve hits the y-axis)." credits to http://www.enchantedlearning.com/history/us/monuments/stlouisarch/

The parablola of the St.Louis Arch opens down, and has a vertex of (315, 630), which is a maximum. The vertex in this situation is the maximum hight the St.Louis Arch reaches. I found the vertex by graphing the parabolic shape of the arch. Since the arch is spans 630 feet as width, the x-axis has to span at least to 630 or more. Since the max height of the parabola is also 630, the y-axis would have to be at least 630 or more tall. The St.Louis Arch does have 60 foot deep foundations keeping the structure stable, so as a negative factor, the parabola will barely cross over the x-axis, which represents the gound, since the foundations are burried underground. The equation for the axis of symmetry for the St.Louis Arch is X=315

## Equations:

Vertex Form: y-630= -0.0063(x-315)Factored Form: y= -0.0063(x-0)(x-630)

Standard Form: y= -0.0063x^2+4x+0

Using the program Quadratic Regression Applet, I was able to figure out the standard for equation for the St.Louis Arch on a graph.

I plugged in all of the numbers and figured out the other equations by figuring out what the other variables stood for.

The y-intercept of my graph would be (0,-60) but the graph from EXCEL is derpy... The y-intercept is (0,-60) because i set the graph so the on leg of the arch would represent the y-axis. It is negative because as stated before, the arch has 60 foot foundations undergound, when the ground is considered the x-axis. I basically found it by graphing the parabola.

## Discriminant:

16-4(-0.0063)x0= 16Since the answer is positive,it means the parabola crosses the x-axis twice, that it is factorable, and since 16 is a perfect square, the two solutions are rational numbers.

## X-Intercepts:

My parabola has two x-intercepts because the x-axis would represent gound level in my graph, and the arch has 60 foot deep foundations underground. They are (0,0) and (0,630) I found them by graphing the standard form equation on abettercalculator.com but you can find them without a graph by using the vertex form(y-630= -0.0063(x-315) ) for the vertex, the factored form (y= -0.0063(x-0)(x-630) ) for one x-intercept, and the standard form (y= -0.0063x^2+4x+0) to find the other x-intercept. The x-intercepts represent when the legs of the arch hit ground level. Factoring the parabola would give you two binomials that would simplify into the standard form equation of the parabola which then can be revealed to the other forms of equations which will eventually give you the x-intercepts.

## QUESTION!!!

Inside the arch are a series of spherical elevators that take you from 60 feet below the arch, to nearly the top of the arch where the observatory is. We want to find out how long it would take (in seconds) for the elevator to be 420 feet from the starting point. When will the elevator be 420 feet from the starting point? (hint, two solutions)Since we are looking for when the elevator is 420 feet from the starting point, y=420

Fit in with the standard for equation for the arch, the equation is 420= -0.0063x^2+4x+0

In order to put this into a quadratic equation, we need 420 to be on the other side, so subtract. y=-0.0063^2+4x-420

Now we have our

a, bandcnumbers, we can plug it all into the quadratic equation.## PARABOLAAAAAAAAAAAAAAAAA!!!!!!