Polynomial Rollercoaster Project
Mrs. Chernowski
Pre-Calculus
Chris Murphy
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.
Real roots:
1.25 (1 ¼) → (x - 1.25), 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)
Factoring the Polynomial
• Since the visible roots on the graph are: - (real roots and double root)
o 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
o Therefore, (x2-2x+5) = #2.) [x = 1 + 2i, 1 – 2i]
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)
 (x4-10.75x3+40.375x2-62.625x+33.75)(x2-2x+5)
 x6-12.75x5+66.875x4-197.125x3+360.875x2-380.625x+168.75
#.3)
Complete factored polynomial:
y = h(t) = x6-12.75x5+66.875x4-197.125x3+360.875x2-380.625x+168.75
• This polynomial function depicts the rollercoaster with a leading coefficient of
1, although it is not in standard form because it currently includes decimals.
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:
o y = h(t) = x6-12.75x5+66.875x4-197.125x3+360.875x2-380.625x+168.75
 The decimals of the coefficients and constant are then transformed into fractions to
obtain the LCD (Least Common Denominator)
o x6-12(¾)x5+66(⅞)x4-197(⅛)x3+360(⅞)x2-380(⅝)x+168(¾)
o Since the LCD is 8, the function is multiplied as a whole by 8 to eliminate the
fractions.
o 8 * [x6-12(¾)x5+66(⅞)x4-197(⅛)x3+360(⅞)x2-380(⅝)x+168(¾)]
o 8x6-102x5+535x4-1577x3+2887x2-3045x+1350
#4.)
Polynomial in standard form:
h(t) = 8x6-102x5+535x4-1577x3+2887x2-3045x+1350

#.5) To
verify the roots of the equation, synthetic division is applied to all roots:
o Roots: 1.25, 2, 3, 4.5, 1+2i, 1-2i
1.25|
8
-102
535
-1577
2887
-3045
1350
↓
Since the remainder
10
-115
525
-1315
1965
-1350
is 0, (x-1.25) is a
8
-86
363
-851
1185
-675
|
0
factor of the entire
-------------------------------------------------------------------------------------------------------------------
÷
polynomial.
2|
÷
8
-102
535
-1577
2887
-3045
1350
↓
16
-172
726
-1702
2370
-1350
8
-86
363
-851
1185
-675
|
0
-------------------------------------------------------------------------------------------------------------------
3|
8
-102
535
-1577
2887
-3045
1350
÷
↓
4.5|
8
-102
535
-1577
2887
-3045
1350
÷
↓
36
-66
-297
238
1071
-506
-2277
610
2754
-300
-1350
|
0
24
-234
903
-2022
2595
-1350
8
-78
301
-674
865
-450
|
0
-------------------------------------------------------------------------------------------------------------------
8

Now the imaginary root(s) must be synthetically
divided to verify it as a factor of the polynomial.
#.5)
o 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).
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:
oy
approaches -∞ as x approaches (+)∞
 But there is a range and domain restriction that does not
make this behavior continuous.
#8.)
Practical domain:
 0 ≤ x ≤ 4.5
[0, 4.5]
#9.)
Practical range:
 -50 ≤ y ≤ 350
[-50, 350]
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)