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Math

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Mimi Luo

on 20 March 2015

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Transcript of Math

Functions
Mimi Luo & Emma Swift
Tower of Pisa
Louvre Pyramid
Eiffel Tower
Roman Aqueducts
London Eye
Dimensions
Radius: 60 m
Full height: 135 m

Parent:
(x-h)^2 + (y-k)^2 = r^2

Max Temperatures in Germany
Dimensions
Height: 21.6 meters
Square base: 35 meters
The height of the pyramid represents the highest y-coordinate the line can reach. This translates to the peak of the equation.
In order to be centered, the base must be divided in half, resulting in x-intercepts of -17.5 and 17.5.
Parent:
IxI=f(x)
Dimensions
Height: 301 meters
Width of base: 125 meters
The base of the Eiffel Tower is a square, about 125 meters in length. An exponential function would only show one side of the Eiffel Tower and the base would therefore need to be cut in half.
125/2= 62.5
This would represent the x-intercept. An exponential function will never have a x-intercept. Because of this, the closest coordinate one can get is 0.42.
Parent:
10^x=f(x)
Dimensions
Taller peak: 5007 m
Shorter peak: 4003 m
Trough: 4000 m
Horizontal distance between peaks: 2490 m
Width at sea level: 5834 m

Parent:
f(x)=x^4

Dimensions
Height (on low side): 55.86 meters
Displacement: 3.9 meters
The Tower leans at a 3.99° angle. This means that the top of the Tower is displaced by 3.9 meters horizontally.
If the y-intercept is 55.86, then the x-intercept would have to be -3.9, because that is the number of meters that it is displaced.
Parent:
f(x)=ax+b
Swiss Alps
f(x) = (5.57 * 10^-12) (x - 2126) (x + 3708) x^2 + 4000
Dimensions
Height: 1.8 meters
Width (on bottom): 5 meters
Because the y-intercept is 1.8, the highest point, the x-intercepts must be half of the width.
5/2= 2.5
Because a parabola will intercept the x-intercept at two places and the peak would be in the middle, the two x-intercepts would be -2.5 and 2.5.
Because both intercepts are none, the zeroes may be used to solve for "a". I used the coordinate (2.5,0) in order to solve for "a".
a(2.5)^2+1.8=f(0)
a=-0.288
Parent:
ax^2+b=f(x)
I placed the y intercept at the point of the trough, so the points would be (-3708,0) ,(-2220,5007) (0,4000), (270,4003), (2126,0)
Because I know the zeroes of the equation, along with the parent function and y intercept, I formed the following equation
f(x) = a (x - 2126) (x + 3708) x^2 + 4000
I then utilized a known data point, (270,4003), and plugged both into the equation to solve for a.
f(270) = a (270 - 2126) (270 + 3708) 270^2 + 4000 = 4003
a= -5.5737948*10^-12
So the equation is
f(x) = (-5.57 * 10^-12) (x - 2126) (x + 3708) x^2 + 4000
-.288x^2+1.8=f(x)
301*.9^x=f(x)
-1.234*IxI+21.6=f(x)
I know that it is held up 15 meters in the air because the total height minus the diameter is 15 (135-120=15).
Therefore, the center of the circle is 15(how much it is held up)+60(the radius) = 75 m above the origin. Thus, the center is (0,75). The radius is 60, and 60^2 = 3600. I inserted each into the parent function to reach
x^2 + (y-75)^2 = 3600
f(x)=14.32x+55.86
x^2 + (y-75)^2 = 3600
London Eye
Dimensions:
Peak (7pi/4, 24)
Trough (pi/4, 2)
Amplitude: 11
Period: 3pi
Vertical Shift: 13
Phase shift: 7pi/8

Parent :
f(x)=a sin (bx-c) + d
I changed each month to be pi/4 so that I could create a trigonometric function. To get the vertical shift, I added the amplitude to the lowest f(x) value, which gives the location of the midpoint, 13 (11+2=13). The midpoint of the parent is 0, so the vertical shift is 13.
The parent function usually begins at 0, the midpoint, and in this case the function does not start at the midpoint, 13. Instead, it begins at 0, therefore there must be a phase shift. (0,0) to (7pi/8, 13) is a horizontal shift of 7pi/8.
I then inserted the known information into the parent function and got f(x)=11(sin2/3(x-7pi/8))+1
f(x)=11(sin2/3(x-7pi/8))+13
Auschwitz
Dimensions:
Width: 3.7m
Height (from the curve to the top of the arch): .1 m
Parent:
f(x)=log(x)
I placed the curve of the graph [which is (1,0) on the parent] at the origin.
Since the width is 3.7m and the height (from the curve to the tip of the arch) is halfway through the doorway, I know one of the points is (1.85, .1) because 3.7/2=1.85. With the knowledge of this and the fact that the curve is at the origin, I can deduce that the equation is f(x)=a log(x+1). When I input a known point, such as (1.85, .1), and solve for a, I find the equation
f(x)= .2199log(x+1)
h
f(x)= .2199log(x+1)
References
Aqueduct of Segovia. (n.d.). Retrieved March 20, 2015, from http://en.wikipedia.org/wiki/Aqueduct_of_Segovia
Britain Express. (n.d.). London Eye [Photograph]. Retrieved from http://www.britainexpress.com/zen/albums/potd/London-Eye-0371.jpg
Crystal Clear Maths. (2013, March 27). How to Construct a Polynomial Function Given Its Graph [Video file]. Retrieved from
Desmos graphing calculator. (n.d.). Retrieved March 19, 2015, from Desmos website: https://www.desmos.com/calculator
Encyclopaedia Britannica. (2006, June 28). Eiffel Tower [Photograph]. Retrieved from http://media-2.web.britannica.com/eb-media/67/114967-004-E3A58022.jpg
Hardy Services. (n.d.). Leaning Tower of Pisa [Photograph]. Retrieved from http://hardyservices.co.uk/wp-content/uploads/2014/03/PIZA.jpg
Holocaust concentration camps [Photograph]. (2014, February 1). Retrieved from http://cdn.history.com/sites/2/2014/01/auschwitz-entrance-P.jpeg
Lambeth. (2000). London Eye. Retrieved March 19, 2015, from Structurae website: http://structurae.net/structures/london-eye
Leaning Tower of Pisa Facts. (n.d.). Retrieved March 20, 2015, from http://www.towerofpisa.info/Tower-of-Pisa-facts.html
Leaning Tower of Pisa. (n.d.). Retrieved March 20, 2015, from http://en.wikipedia.org/wiki/Leaning_Tower_of_Pisa
Roman aqueducts [Photograph]. (n.d.). Retrieved from https://encrypted-tbn1.gstatic.com/images?q=tbn:ANd9GcQqqR--Vos6JtouDUB5yA3voKC8ndMfVtUM7TyT3H3yUzYM5O88cw
School German language trips to the Rhineland, Germany. (n.d.). Retrieved March 19, 2015, from Rayburntours website: http://www.rayburntours.com/education/language/the-rhineland/
Wikipedia. (n.d.). Louvre Wikipedia [Photograph]. Retrieved from https://mahasiswaarsitektur.files.wordpress.com/2013/03/louvre-bannenhaff-mat-pyramid-w.jpg
Zehring, M. (2008, May 13). Finsteraarhorn with helicopter [Photograph]. Retrieved from http://images.summitpost.org/original/404345.jpg

Absolute Value Function
Exponential Function
Conic Section
Polynomial Function
Quadratic Function
Linear Function
Sine Function
Logarithmic Function
Full transcript