1. Parabolas in Real Life

2. The Eiffel Tower 3 Interesting facts:
The Eiffel tower is 986 feet tall and is constructed out of iron material.
The Eiffel Tower was built in 1889 and was the tallest structure in the world until 1930.
The tower was named after its designer and engineer, Gustave Eiffel, and over 5.5 million people visit the tower every year.

3. We can see on the graph that the roots of the quadratic are:
x = −32.12 (since the graph cuts the x-axis at x = − 37.12);
and
x = 32.12 (since the graph cuts the x-axis at x = )
Now, we can write our function for the quadratic as follows (since if we solve the following for 0, we'll get our 2 intersection points):
f(x) = (x  )(x − 37.12)
We can expand this to give: f(x) = x2  − 1,
This is a quadratic function which passes through the x-axis at the required points. But is this the correct answer?
Observe the graph passes through −79.3 on the y-axis. Let's substitute x = 0 into the equation I just got to check if it's correct.
f(0) = 02  − 1, = − 1,
It's not correct!
It turns out there are an infinite number of parabolas passing through the points (−37.12,0) and (37.12,0).
We need a different method…
(0, 36.9)
(-37.12, 0)
(37.12, 0)
Scale: 1 = 3.712m

4. Scale: 1 = 3.712m Vertex Form of a Quadratic Function
We can see the vertex is at (-2, 1) and the y-intercept is at (0, 2).
We can see the vertex is at (0, 79.3) and the y-intercept is at (0, 79.3).
We just substitute as before into the vertex form of our quadratic function.
We have (h, k) = (-2, 1) and at x = 0, y = 2.
(h,k) = (0,79.3), and x = 0, y = 79.3 So
y = a(x − h)2 + k
becomes
79.3 = a(0 − 0)2 + 79.3
79.3 = 79.3
Which is not untrue, but it is not the value of a
Then the problem should be the parabola placement on the Cartesian Plane…
(0,36.9)
(-37.12, 0)
(37.12, 0)
Scale: 1 = 3.712m

5. Vertex Form of a Quadratic Function
We can see the vertex is at (37.12, 36.9) and the y-intercept is at (0, 0).
We just substitute into the vertex form of our quadratic function.
We have (h,k) = (37.12,36.9), and x = 0, y = 0
So, the vertex formula of a quadratic function
y = a(x − h)2 + k
Becomes
0 = a(0 − 37.12)2 + 36.9
0 = 1, a
= 1, a
1, ,
= a
So our quadratic function is:
f(x) = (x − 37.12)2 + 36.9
f(x) = (x2 - 74.24x + 1, )
f(x) = x2 + 1.9881x 
f(x) = x2 + 1.9881x 
(37.12,36.9)
(0, 0)
(74.24, 0)
Scale: 1 = 3.712m

6. Test on a 3 points on the graph:
f(x) = x2 + 1.9881x 
Equation for Line of Symmetry
x =
2 ( )
x ≈ 37.12
Calculated Vertex Coordinates
f(37.12) ≈ (37.12)2 + 1.9881(37.12)
f(37.12) ≈ ≈
Coordinates of Vertex: (37.12,36.9) x y
11.16
18.85
22.27
30.99
66.82
13.29
f(11.16) = (11.16)2 + 1.9881(11.16) 
f(22.27) = (22.27)2 + 1.9881(22.27)
f(66.82) = (66.82)2 + 1.9881(66.82)
(37.12,36.9)
(18.72, )
(9.36, 18.56)
(66.82, 13.29)
(0, 0)
(74.24, 0)
Scale: 1 = 3.712m

7. Resources