y=-0.1458333333x2+0.0416666667x+13.5
y=-0.1458333333(x-0.1426611797)2+13.5
This is the St. Louis Arch, also called the Gateway Arch or
the Gateway to the West, is the main attraction of the Jefferson National
Expansion Memorial in St. Louis ,
Missouri . It’s construction
started in February 12, 1963, and was finished in October 28, 1965. The Gateway
Arch weighs 17,246 pounds and used 900 tons of stainless steel. I chose it
because it was the first thing that popped into my head when I though of
parabolas in real life, and I thought I it would be interesting to find the
equation of one of USA ’s famous monuments.
(-4, 11), (0, 13.5), (2, 13), (-6, 8), (6, 8.5), (12, -7),
(14, -14.5), (18, -33), (-10, -1.5), (-12, -8)
This graph I used the three points (0, 13.5), (2, 13), (12,
-7) for the points. I made a matrix of it (0, 0, 1, 13.5), (4, 2, 1, 13), (144,
12, 1, -7) and used the rref function on my graphing calculator and got
a=-0.1458333333, b=0.0416666667, and c=13.5. I plugged those numbers into my
graphing to graph the line into there and then compared whether or not they
looked alike. Satisfied, I went and plugged the equation into Holt’s Graphing
Calculator to find the points on the line.
This is the bottom view of the Eiffel Tower .
Located in Paris , France , the construction was
started in January 28, 1887, and was finished in March 31, 1889. The Eiffel Tower
uses 7.5 kilowatts of electricity annually to light up the tower at night. I
chose the Eiffel Tower
because not many people think of parabolas when they see the Eiffel Tower .
Most people see the pyramid-like shape, but when I searched up “parabolas in
real life”, this picture showed up, so I though it would be pretty creative to
find the equation of the parabola of the Eiffel Tower .
y=-0.0625x2+2
y=-0.0625x2+2
(0, 2), (4, 1), (-4, 1), (8, -2), (-8, -2), (12, -7), (-12,
-7)
I found out the equation of this graph by using the three
points (0, 2), (4, 1), and (8, -2). I put the numbers into a matrix ( 0, 0, 1,
2) (16, 4, 1, 1) (64, 8, 1, -2) and I used rref on my graphing calculator to
find that a=-0.0625, b=0, and c=2. I plugged those numbers into my graphing to
graph the line into there and then compared whether or not they looked alike. Satisfied,
I went and plugged the equation into Holt’s Graphing Calculator to find the
points on the line.