Parabolas in Real Life The Eiffel Tower 3

Parabolas in Real Life

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.

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 = 37. 12. ) 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 + 37. 12)(x − 37. 12) We can expand this to give: f(x) = x 2 − 1, 377. 8944 This is a quadratic function which passes through the x-axis at the required points. But is 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, 377. 8944 = − 1, 377. 8944 It's not correct! It turns out there an infinite number of parabolas passing through the points (− 37. 12, 0) and (37. 12, 0). (0, 36. 9) We need a different method… (-37. 12, 0) (37. 12, 0) Scale: 1 = 3. 712 m

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 yintercept 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 Which is not untrue, but it is not the value of a (0, 36. 9) Then the problem should be the parabola placement on the Cartesian Plane… (-37. 12, 0) (37. 12, 0) Scale: 1 = 3. 712 m

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 Becomes y = a(x − h)2 + k 0 = a(0 − 37. 12)2 + 36. 9 0 = 1, 377. 8944 a + 36. 9 -36. 9 -36. 9 = 1, 377. 8944 a 1, 377. 8944 -0. 02678= a So our quadratic function is: (37. 12, 36. 9) f(x) = -0. 02678(x − 37. 12)2 + 36. 9 f(x) = -0. 02678(x 2 - 74. 24 x + 1, 377. 8944) + 36. 9 f(x) = -0. 02678 x 2 + 1. 9881 x - 36. 9 + 36. 9 f(x) = -0. 02678 x 2 + 1. 9881 x (0, 0) (74. 24, 0) Scale: 1 = 3. 712 m

Test on a 3 points on the graph: f(x) = -0. 02678 x 2 + 1. 9881 x x y 11. 16 18. 85 f(11. 16) = -0. 02678(11. 16)2 + 1. 9881(11. 16) 22. 27 30. 99 f(22. 27) = -0. 02678(22. 27)2 + 1. 9881(22. 27) 66. 82 13. 29 f(66. 82) = -0. 02678(66. 82)2 + 1. 9881(66. 82) Equation for Line of Symmetry x = -1. 9881 2 (-0. 02678) x ≈ 37. 12 (37. 12, 36. 9) Calculated Vertex Coordinates (18. 72, 22. 28 ) (9. 36, 18. 56) (66. 82, 13. 29) (0, 0) (74. 24, 0) Scale: 1 = 3. 712 m f(37. 12) ≈ - -0. 02678(37. 12) 2 + 1. 9881(37. 12) f(37. 12) ≈ - 36. 8982 + 73. 7982 ≈ 36. 9 Coordinates of Vertex: (37. 12, 36. 9)

Resources • https: //upload. wikimedia. org/wikipedia/commons/3/3 c/Eiffel _plan. jpg • https: //commons. wikimedia. org/wiki/Category: Architectural_ drawings_of_the_Eiffel_Tower#/media/File: Dimensions_tour _Eiffel. JPG • https: //commons. wikimedia. org/wiki/Category: Architectural_ drawings_of_the_Eiffel_Tower#/media/File: Skizze_Eiffelturm _-_technische_Daten. png