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Polynomial Rollercoaster Project

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1 Polynomial Rollercoaster Project
Mrs. Chernowski Pre-Calculus Chris Murphy

2 Key: t (x-cord) in terms of minutes. h(t) (y-cord) in terms of feet.
Requirements: At least 3 relative maxima and/or minima The ride length must be at least 4 minutes The coaster ride starts at 250 feet The ride dives below the ground into a tunnel at least once Key: t (x-cord) in terms of minutes. h(t) (y-cord) in terms of feet.

3 Real roots: 1.25 (1 ¼) → (x ), 2 → (x - 2), 3 → (x - 3), 4.5 (4 ½) → (x - 4.5) Double root: - (When the graph touches the x-axis once producing the same root twice.) 3 → (x - 3)2 Imaginary root: 1 + 2i, 1 - 2i → (x2 – 2x + 5)

4 Factoring the Polynomial
Since the visible roots on the graph are: - (real roots and double root) 1.25, 2, 3, and 4.5 this concludes to → #2.)x= 1.25, 2, 3, 4.5 Since the polynomial also requires an imaginary root I included a function that results in the inclusion of “i” (an imaginary number) by using a quadratic equation that cannot be simplified into a real number by the quadratic formula. This imaginary root was assured by selecting values that allow a negative value in the “square root section”: Selected example quadratic equation: (x2-2x+5) Selected quadratic formula values: a = 1, b = -2, c = 5 x= [-(-2) ± √(-2)2 – 4(1)(5)]/2 → (2 ± √-16)/2 → “1 ± 4i” (i = √-1) x = 1 + 2i, 1 – 2i Therefore, (x2-2x+5) = #2.) [x = 1 + 2i, 1 – 2i]

5 Factoring the Polynomial (continued)
In conclusion, all of the roots (real, double, and imaginary) can be stated as: (x – 1 .25)(x – 2)(x – 3)(x – 4.5)(x2 – 2x + 5) (x2-2x-1.25x+2.5)(x2-4.5x-3x+13.5)(x2-2x+5) (x2-3.25x+2.5)(x2-7.5x+13.5)(x2-2x+5) (x x x x+33.75)(x2-2x+5) x x x x x x #.3) Complete factored polynomial: y = h(t) = x x x x x x This polynomial function depicts the rollercoaster with a leading coefficient of 1, although it is not in standard form because it currently includes decimals.

6 Factoring the Polynomial Standard Form
Standard form: - The coefficients of each degree (including the constant) must be represented as integers (a number that does not contain decimals/fractions) Since the polynomial equation was distinguished as: y = h(t) = x x x x x x The decimals of the coefficients and constant are then transformed into fractions to obtain the LCD (Least Common Denominator) x6-12(¾)x5+66(⅞)x4-197(⅛)x3+360(⅞)x2-380(⅝)x+168(¾) Since the LCD is 8, the function is multiplied as a whole by 8 to eliminate the fractions. 8 * [x6-12(¾)x5+66(⅞)x4-197(⅛)x3+360(⅞)x2-380(⅝)x+168(¾)] 8x6-102x5+535x4-1577x3+2887x2-3045x+1350 #4.) Polynomial in standard form: h(t) = 8x6-102x5+535x4-1577x3+2887x2-3045x+1350

7 Dividing the Polynomial (Synthetic Division)
#.5) To verify the roots of the equation, synthetic division is applied to all roots: Roots: 1.25, 2, 3, 4.5, 1+2i, 1-2i 1.25| ÷ ↓ | 2| ÷ ↓ 3| ÷ ↓ | 4.5| ÷ ↓ | Since the remainder is 0, (x-1.25) is a factor of the entire polynomial.

8 Dividing the Polynomial (Synthetic Division) (continued)
#.5) Now the imaginary root(s) must be synthetically divided to verify it as a factor of the polynomial. Imaginary roots: 1-2i, 1+2i → “(x2 – 2x + 5)” The depressed polynomial (the coefficients above) from 1-2i is then used as the dividend when using 1+2i as the divisor (as shown below).

9 End Behavior of Function
Since the leading coefficient An has an even degree “6” and An > 0 the graph will rise to both the left and the right. At the decreasing interval at the end of the graph represents: y approaches -∞ as x approaches (+)∞ But there is a range and domain restriction that does not make this behavior continuous.

10 The Practical Range and Domain of the Graph
#8.) Practical domain: 0 ≤ x ≤ 4.5 [0, 4.5] #9.) Practical range: -50 ≤ y ≤ 350 [-50, 350]

11 Increasing, Decreasing, and Constant Intervals
#10.) Increasing intervals: (0 ≤ x ≤ 0.5), (1.5 ≤ x ≤ 2.5), (3 ≤ x ≤ 3.5) → (0, 0.5)U(1.5, 2.5)U(3,3.5) Decreasing intervals: (0.5 ≤ x ≤ 1.5), (2.5 ≤ x ≤ 3), (3.75 ≤ x ≤ 4.5) → (0.5, 1.5)U(2.5, 3)U(3.75, 4.5) Constant interval: (3.5 ≤ x ≤ 3.75) → (3.5, 3.75)


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