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Math Project 2

Math Project 2. By: Holden Greene and Joseph Young. Part 1. We plugged the numbers from the table into the formula that was given. h=at^2+bt+c 1) 23645=4a+2b+c 2) 32015=400a+20b+c 3) 33715=1600a+40b+c. Part 2.

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Math Project 2

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  1. Math Project 2 • By: Holden Greene and Joseph Young

  2. Part 1 • We plugged the numbers from the table into the formula that was given. h=at^2+bt+c • 1) 23645=4a+2b+c • 2) 32015=400a+20b+c • 3) 33715=1600a+40b+c

  3. Part 2 • First we subtracted problem 1 from problem 2. Using Elimination to get the answer. • 32015=400a+20b+c • - 23645=-4a-2b-c • ---------------------------- • 8370=396a-18b • This answer becomes #4.

  4. Part 2 Continued • We then subtracted problem 2 from 3 and used elimination to solve it again. • 33715=1600a+40b+c • -32015=-400a-20b-c • --------------------------- • 1700=1200a+20b • This answer becomes problem 5.

  5. Part 2 Continued • We then used elimination with problems 4 and 5. We multiplied 4 by 10 and 5 by -9 in order to eliminate. • 8370=396a+18b (10)= 83700=3960a+180b • 1700=1200a+20b (-9)= -15300=-10800a-180b • ----------------------------------------------------------- • 68400=-6840a • then divide by -6840 which gives you an answer of -10=a

  6. Part 2 Continued • We then plugged the A(-10) that we got from our earlier solution into problem 5. • 1700=1200(-10)+20b • 1700=-12000+20b • 13700=20b • _____________________ • The answer that we got was 685=b.

  7. Part 2 Continued • We then plugged our answers for A(-10) and B(685) into our first problem to find C. • 23645=4(-10)+2(685)+C • 23645=-40+1370+C • 23645+40-1370=C • _____________________ • The answer that we got was C=22315.

  8. Part 2 Summary • In summation of our work the answers that we got were: • A=-10 • B=685 • C=22315

  9. Part 3 • We then plugged our numbers into our equation h=at^2+bt+c. • This gave us h=-10t^2+685t+22315

  10. Part 4 • We found the maximum value of the quadratic function by solving for Y from our equation in Part 3. We are then able to come up with our maximum

  11. The Solving of Part 4 • y=-10t^2+685t+22315 • y=-10(t^2-68.5t+1173.0625)+22315+-10(1173.0625) • 22315+11730.625 • y=-10(t-34.25)^2 • y=-10(t-34.25)^2+34045.625 • This gives us our graph and the parabola that we will be using for our answer. A=-10 X=T H=-34.25 K=34045.625

  12. Part 5, The Graph • We used this equation to get our graph y=-10(t-34.25)^2+34045.625. • Maximum= 34045.625 • Vertex=(34.25, 34045.625) • Axis X= 34.25

  13. Part Five, The Graph Continued

  14. Part 6 • Yes it changed our minds completely about the purpose of math and the way that it can be used in real world situations. It really enlightened our thinking towards the real world examples that math participates in and give credence towards math and the purpose that it serves in the world outside of the boundaries of our educational facilities. Math is an ever important factor in life more so probably then English or a history or social studies class of some sort. This is due to the ever present factor of math in our life while rarely must we remember history or the role that it plays in our world. And english is obviously a very unneeded skill that will serve no purpose later in life, because very rarely do we use english in our lives. That is why math is very important to us and is how this project has revealed to us the importance of math in the real world. We use math alot more then any other subject.

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