Question

As one of the new roller coaster engineers, you have been tasked with developing a roller coaster that will intertwine with existing Oakville Lake Amusement Park structures. For one of the more thrilling sections, the roller coaster will dive down in-between buildings, plummet underground, pop back up, and coast over a hill before shooting back underground. There must be three distinct points where the roller coaster crosses the x–axis. Precise measurements and attention to detail are very important.

Answer 1

First, here is the existing map of current structures. It is important that the roller coaster does not go through the foundation of any of these structures.

1st point: ___6___
2nd point:___-2___
3rd point: ___-7___

Using the points above as zeros, construct the polynomial function, f(x), that will be the path of your roller coaster. Show all of your work.


Using both fundamental Theorem and Descartes` rule of signs, prove to the construction foreman that your funtion matches your graph. Use complete sentences.


Solve for the y intercept for your function, f(x), and then construct a rough graph of your roller coaster. If your y intercept is off the graph, give the coordinates of the y intercept.

Answer 2

The 3 points i have where it crosses the x-axis is -5, 4, and -13

Answer 3

**5, -4, -13

Answer 4

So what are your three points here?
"1st point: ___6___
2nd point:___-2___
3rd point: ___-7___
"

Cause points come in the form (x,y) so idk if you mean that those are the zeros or if 5, 4 and -13 are the zeros

Answer 5

According to the question the
"1st point: ___6___
2nd point:___-2___
3rd point: ___-7___
"
are my zero's.

It also says that the rollercoaster has to pass through the x axis 3 times and the points i chose is 5, -4, -13

Answer 6

Ok, that makes sense. Ok, you gotta see that the zeros of a function are also called the roots of a function. They tell you when the function pass through the x-axis.

Answer 7

So your graph would pass through the x-axis at (6,0) (-2,0) and (-7,0)

Answer 8

Okay so then how would i set up the function for the path that the rollercoaster goes through

Answer 9

@doulikepiecauseidont

Answer 10

Alright, well I might as well explain this as well since you need to include this in your answer. There's a theorem called the Fundamental theorem of algebra. You don't need the specifics for this question but in basics, it says that

"Any polynomial of degree n ... has n roots"
It also follows that it has n roots.
This picture should help explain what I mean, after you look it over tell me and then we can proceed.

Answer 11

sooo would the equation be: y=(x+6)(x-2)(x-7)

Answer 12

Ahh, I see you've gotten tricked. The factors would go like this \[\Huge (x-(6))\rightarrow (x-6)\]
\[\Huge (x-(-2))\rightarrow(x+2)\]
\[\Huge (x-(-7))\rightarrow(x+7)\]

Answer 13

The formula is (x-a)(x-b)(x-c)... where a,b,c are the factors that they gave you. So subtracting a negative number (-2 and the -7) gives you x+ that number

Answer 14

Oh okay, so then the equation would actually look like y=(x-6)(x+2)(x+7)

Answer 15

Correct! Now that is your graph.

Answer 16

no using the fundamental Theorem and Descartes` rule of signs how would i prove the function matches the graph

Answer 17

now**

Answer 18

This is the graph btw in case you have to draw it.
https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=(x-6)(x%2B2)(x%2B7)

Answer 19

Well we have shown that using that the fundamental theorem that this polynomial if we are given the roots of the polynomial (-7, -2, and 6) all we have to do is use the equation (a(x-r1)(x-r2)(x-r3)...) where r1,r2...are the roots to get our expression.

Answer 20

The expanded expression is \[\Large x^3-12x^2 + 41x - 30\]

Answer 21

Okay and does this equation need factoring or would it work as is

Answer 22

Well we know what looks like factored (x-6)(x+2)(x-7)=\[\Large x^3-12x^2 + 41x - 30\]

Answer 23

lol good point, so now for the last question, how would i find the y intercept

Answer 24

Now we need to use the Descarte Rules of Sign using the expanded function. Sorry I was looking it over and here it is for further reding but I'll explain it really fast
http://www.purplemath.com/modules/drofsign.htm

Answer 25

The Rule of Signs looks directly at the sign of the numbers (positive or negative number) of the function and count how many times it changes

Answer 26

okay... im a little lost, are we working on 2 or 3?? sorry

Answer 27

2 still, We're working on explaining, using Descartes's Rules of signs that our function matches what we wrote, we already used the FTA. Anyway, we need to only worry about the sign in front of the numbers not what they are.

Answer 28

Okay...so im a little confused on the concept of Dascartes's Rules of sign's. How exactly would you explain how it proves the equation..

Answer 29

We find that there are 3 sign changes (it changes from positive to negative to positive back to negative), right?

Answer 30

yes

Answer 31

Also, the website also says "By the way, in case you're wondering why Descartes' Rule of Signs works, don't. The proof is long and involved;"

Answer 32

This only tells you about the nature of the roots, not how to find them or what they are. It just tells you, as you're about to see if you have x number of positive/negative roots. Back to my explanation, we have just done the "positive case".

The positive case is used when the first term (in our case is x^3) is positive and tells you how many roots you MAY have, based on the sign changes.

Notice the word may. When we find the number of sign changes we then conclude that this function can have that many positive roots, or anything 2 less than that. So for example, we found that we have 3 sing changes. So that means we can conclude our function has 3 or 1 positive roots.

It's actually only 1 (the point (6,0)) but that's just how this rule works. Do you follow, so in your paper you can say "there are three sign changes so that means we have either 3 or 1 positive roots, in this case we only have one positive root and that is 6"

Answer 33

Okay i am following thank you so much for helping me out. and lastly for #3 how would i find the y intercept

Answer 34

We then would have to proceed with the "negative case" to find out if there any negative roots (we have two and they are -2 and -7 as you know).

To do so we have to take f(x) which is our function
\[\Large f(x)= x^3-12x^2 + 41x - 30\]

And we have to find the sign changes when we make everything negative, in other words we have to find the sign changes in f(-x). Do you know how to go from f(x) to f(-x)?

Answer 35

Im not 100% sure but wouldnt you divide each side by -1

Answer 36

No, you would replace, everytime you see an x, with -x, just like if you were to solve this for f(2) you would do \[\Large f(2)=(2)^3-12(2)^2+41(2)-30\]

Answer 37

This would then give you \[\Large f(-x)=(-x)^3-12(-x)^2+41(-x)-30\]

Answer 38

Okay i see what you mean so now the equation should look like ... well nevermind you just put it xD

Answer 39

A hint: a negative number raised to an odd number is negative so like (-2)^3=-8
a negative number raised to an even power is positive (-2)^2=4

This helps simplify this to

\[\Large f(-x)=-x^3-12x^2-41x-30\]

Answer 40

the x^3 and x are the only ones that switch since thwy have odd powers (3 and 1) and the -12(-x)^2 just simplifies to -12x^2

Anyways, you can see there are 0 sign changes. So we can conculude That there are 2 or zero negative roots

2 negative roots? But there weren't two sign changes? Those are good questions. Remember when we used the Fundamental theorem of algebra? Our function has a degree of 3 (the 3 on the x^3 tells us the degree). That means that we have at least 3 roots. So with that we can have any combination of positive and negative roots that's less than or eqaul to three and we can surely have 0 - and 3 + roots as well as 2 - and 1 + root (which is in our case, which is another reason why I'm saying that). So you can show, with all of this that:

"In the positive case, we found that there are either 3 or 1 positive roots and in the negative case, using Descarte's rule of signs that there are 2 or zero negative roots. In our case, there are 2 negative roots and 1 positive roots, and since our function is of degree 3, our polynomial can have at most 2+1=3 roots"