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Gateshead Millennium Bridge

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by

Marissa Melton

on 13 November 2013

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Transcript of Gateshead Millennium Bridge

QUADRATICS PROJECT
“Work of Arch”
By:Marissa Melton per.5

Coordinates into Standard Form
f(x)=-9/845 x^2+18/13 x+0
(130,0)
0=-9/845 (130)^(2 )+18/13 (130)+0
0=-9/845 (16900)+180+0
0= -180+180+0
0=0
f(x)=-9/845 x^2+18/13 x+0
(0,0)
0=-9/845 (0)^(2 )+18/13 (0)+0
0=0
f(x)=-9/845 x^2+18/13 x+0
(65,45)
45=-9/845 (65)^2+18/13 (65)+0
45=-9/845(4225)+90
45=45

Background Information
The Gateshead Millenium Bridge was built across the Tyne River in England. The bridge was designed by Wilkinson Eyre Architects and engineered by Gifford. The bridge is two curves, one forming the deck and the other supporting it, spanning between two islands running parallel to the quaysides. It is a pedestrian and cyclist bridge. It uses a tilting mechanism that allows ships to pass through. Due to this feature and it's shape the bridge is often called the "Winking Eye Bridge".
Vertex Form into Standard Form
f(x)=-9/845 (x-65)^2+45
Plug in the a value and the vertex.
f(x)= -9/845 (x-65)(x-65)+45
Factor the (x-65)^2 into two binomials
f(x)=-9/845 (x^2-130x+4225)+45

f(x)=-9/845 (x〗^2)-9/845 (-130x)-9/845 (4225)+45
Distribute a.
f(x)=-9/845 x^2+ 18/13x +0
Simplify & group like terms. The equation is now in standard form f(x)=ax^2+bx+c.
Gateshead Millennium Bridge
Vertex Form
Vertex Form: f(x)=a〖(x-h)〗^2+k
Vertex (65, 45)
f(x)= a〖(x-65)〗^2+45
0=a(4225)+45
-45= a(4225)
-45/4225=a
-8/(845 )=a
Solve for a value.

0 13 26 39 52 65 78 91 104 117130 x
y
54
45
36
27
18
9

points: (0,0) (65,45) (130,0)


Measurments of bridge
lenght of deck: 130m
height of arch: 45m

axis of symmetry
In vertex form:
f(x)=a(x
-h
)^2+k
f(x)= a(x
-65
)^2+45

x=h


x=65
standard form:
f(x)=ax^2+bx+c
f(x)=-9/845 x^2+18/13 x+0

x=(-b)/2a
x=(-18/13)/(2(-9/845) )
x=65



.
.
.
Coordinates into Vertex Form
f(x)=-9/845 (x-65)^2+45
(130,0)
0=-9/845(130-65)^2+45
0=-9/845(4225)+45
0=-45+45
0=0
f(x)=-9/845 (x-65)^2+45
(0,0)
0=-9/845 (0-65)^2+45
0=-9/845(4225)+45
0=-45+45
0=0
f(x)=-9/845 (x-65)^2+45
(65,45)
45=-9/845 (65-65)^2+45
45=-9/845(0)+45
45=45








Maximum
The parabala is faced downward and the a value is negative therefore it will have a maximum. The maximum is the vertex (65,45). It is the (h,k) values in the vertex form.
Works Cited:
www.newcastlegateshead.com/...to.../gateshead-millennium-bridge-p238.
http://ribastirlingprize.architecture.com/gateshead-millennium-bridge-2002/
http://www.thw.coventry.sch.uk/Geography/GatesHead.htm
Full transcript