Question

Really need lots of help on this :/
ill give out medals, best answer, and fan
but please help im super behind on math and need to catch up quickly

Answer 1

First, here is the existing map of current structures. It is important that the rollercoaster does not go through the foundation of any of these structures. Pick three points on the x–axis that the rollercoaster will dive underground or resurface. (Points that have a building on them are off-limits.)

-1The rollercoaster will run from right to left, so work in that order when selecting points.
1st point:______
2nd point:______
3rd point:______

Answer 2

-2) Using those points as zeros, construct the polynomial function, f(x), that will be the path of your rollercoaster. Show all of your work.

Answer 3

-3) Using two of the theorems dicussed in the lesson, prove to the construction foreman that your function matches your graph. Use complete sentences.

-4) Solve for the y–intercept for your function, f(x), and then graph your rollercoaster (you can use Geogebra). If your y–intercept is off the graph, give the coordinates of the y–intercept.

Answer 4

@kewlgeek555

Answer 5

Well lets see....let say that we will go underground at x = 6....resurface at x = -2 ...and go underground again at x = -8

Answer 6

this means that our polynomial must have 0's at those points...so we have

\[\large f(x) = (x - 6)(x + 2)(x + 8)\]

*if you were to plug in the points we have...you would get your zeros here...so to find our polynomial function...multiply this out...can you do that?

Answer 7

um yeah let me try

Answer 8

f(x)=(6-6) (-2+2)(-8+8) is that good? im sorry im warning you i realy suck at this :/

Answer 9

Dont get down on yourself :)

alright...so what I meant *which I should have been more clear on* is multiply out the function...

\[\large f(x) = (x-6)(x+2)(x+8)\]

have you used the foil method before? to multiply something like (a + b)(a +b) ??

Answer 10

Nope :/

Answer 11

Isnt it sorta like distributing? :s

Answer 12

Exactly! :) you take the first term in those first parenthesis...and multiply it by each term in the second parenthesis...then do that again for the second term in the first parenthesis.

Answer 13

so we'll start with this...

we have

\[\large f(x) = (x - 6)(x + 2)(x +8)\]

Lets break it up

\[\large (x - 6)(x + 2)\]

can you distribute (foil) this?

Answer 14

Yay ok awesome haha ok give me a sec

Answer 15

would it be x^2+2x-6x-12 ?

Answer 16

Are you in Connections academy? Is this Algebra 2 Foundations b?

Answer 17

Yes c: and im in Algebra 2 idk what Foundations b is :))

Answer 18

@omgkelley hey I'm back...and yes you distributed correctly....just look though...there is a 2x and a -6x....what should we do?

Answer 19

can you help me with Unit three lesson 3?

Answer 20

umm make them smaller? so would it be x & -3 ? @johnweldon1993

Answer 21

do u still need help?

Answer 22

Yessss :c

Answer 23

ok lets see and also did i help in the youtube favor?

Answer 24

I've been stuck ont his for over a week :/

Answer 25

Please help guys :/ If i dont finish this ill most probably fail Math

Answer 26

what @johnweldon1993 was doing is a good start. Let's continue with what he asked you to do with the

2x-6x

So
\[2x-6x=x(2-6)=?\]

Answer 27

umm okay im lost

Answer 28

^_^ no worries

\[2x-6x=-4x\]

Answer 29

All I did above was I pulled the x out in front to make the arithmetic more visible
\[x(2-6)=x(-4)=-4x\]

Can you see where I got that?

Answer 30

um noo i saw that you solved the 2x-6x but you got me lost up there ^

Answer 31

And this was all from the first part of the polynomial he created up top

\[f(x)=(x-6)(x+2)(x+8)\]

We're just looking at the
\[(x-6)(x+2)\]
part first.

\[(x-6)(x+2)=x^2-6x+2x-12
\\ \hspace{2.4cm} =x^2-4x-12\]
So
\[f(x)=\big((x-6)(x+2)\big)(x+8)
\\ \
\\ \qquad = (x^2-6x+2x-12)(x+8)\]

Answer 32

But you understand the

\[2x-6x=-4x\]
if I don't do the "pull x out" thing?

Answer 33

Yes i understand that part :)

Answer 34

then bueno! ^_^

looking at the next part of the distributing then, you have

\[f(x)=\big((x-6)(x+2)\big)(x+8)
\\ \
\\ \qquad = (x^2-4x-12)(x+8)\]

So you just do like you did before, multiplying each of the terms in the quadratic by x, and then again by 8, then simplifying it like did before

Answer 35

oh okay! easy :D

Answer 36

Hey, I'm really sorry, but I have to run to an appointment.

@SolomonZelman Can you help out with the rest of this problem?

Answer 37

oh no D:

Answer 38

In case he's busy, for the last part of the problem you just plug in x=0 into the polynomial you get from simplifying to find the y intercept, so every term with an x in it disappears and you're just left with a number ^_^

Good luck!! You'll be awesome!!!!!!

I'll be back a bit later if you still need help with this ^_^

Answer 39

I probably wiil haha i'll wait up C:
and thanks for being so patient with me haha

Answer 40

Anytime! ^_^

Answer 41

@tester97 have you done this yet?

Answer 42

what exactly is the question?

Answer 43

Theres originally 5 but im doing the alternate assessment which is just 4

Answer 44

oh sorry i didnt do this

Answer 45

Solo is off line..so i will try what i the question?

Answer 46

Okay hold on

Answer 47

Lol there are some confusing things going on in this question...i can't find the question...

Answer 48

The rollercoaster will run from right to left, so work in that order when selecting points.
1st point:__6____
2nd point:___-2___
3rd point:____-8__

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

Answer 49

ok so a function is f(x) = mx + b where m is your slope and b is your y.

Answer 50

LLol yeah ik, sorry ^ those are the first two questions. there are four in total

Answer 51

Lets try and work them out piece by piece i think i know how to do this :3

Answer 52

So what do you think ur function is?

Answer 53

omg okay yay! youre such a life saver . umm i dont know im lost on this whole lesson

Answer 54

Lol i said i think :D lets just remember that :D

Answer 55

Still no one elsee seems to even understand the question haha

Answer 56

lol i don't understand why not...all u need to do is make a function...yet finding the slope...i forgot how to find that...do you remember?

Answer 57

Ummm isnt it y2-y1 divided by x2-x1 ?

Answer 58

oh yes...but ur points are:(1,6)(2,-2)(3,-8)? I believe that is it right?

Answer 59

if so...i has no idea this is quite confusing now lol...Hmmm...

Answer 60

@BangkokGarrett this is very hard can u help me with this?

Answer 61

Well if you have any other points ill take them c:

Answer 62

Maffs told u the answer yet it is like deciphering a futuristic code...

Answer 63

What grade are you in?

Answer 64

Hahaha this assignment is sooo confusing

Answer 65

10th but im taking 11th grade classes

Answer 66

lol im in 9th...

Answer 67

Hey @omgkelley Sorry that our paths didn't cross again to finish the problem and that you had to close the question :/

What @undeadknight26 said about the slope isn't what this question is looking for, because it's a cubic polynomial. Instead of a straight line, it'll look something like this.



@johnweldon1996 helped choose the points well, so that your roller coaster would avoid all of the buildings, and not resurface in any foundation (like the problem wanted).

Looking at the polynomial that you came up with again,
\[f(x)=(x-6)(x+2)(x+8)
\\ \ \ \ =(x^2-4x-12)(x+8)
\\ \ \ \ =(x)(x^2-4x-12)+(8)(x^2-4x-12)\]
(you can finish simplyfying ^_^ Just distribute the x in the first expression, the 8 in the second expression, and add up all the like terms)

you can see that the roots of the equation (where the line of the graph crosses the x axis - aka, the points where the y axis is zero) were developed by defining the ground entry points, (8,0) and (-6,0), and exit point (-2,0) as the roots of a cubic polynomial (a polynomial where the highest power x is raised to is 3 - when you finish simplifying the above term, you'll see that there's an x^3 term). Since you know (by definition) that where the function
\[f(x)=(x-6)(x+2)(x+8)\]
exists at y=0, those will be your roots (shown below). If it helps you visualize it better, you can write the function in terms of y and x.
\[y=(x-6)(x+2)(x+8)\]
It's the same thing.

Now, again knowing that where the roller coaster enters and exits the ground all exist at y=0, all you have to do is set y=0 to prove these are the points you need for your function! Since the ground is at
\[y=0\]
then
\[(x-6)(x+2)(x+8)=0\]
giving the three distinct "roots," or values of x that satisfy the equation (chosen from points on the graph initially). Those are, as chosen,
\[\qquad \begin{array}{l}
x-6=0 & \longrightarrow & x=6 & \rightarrow & \text{corresponds to point (6,0)}
\\x+2=0 & \longrightarrow & x=-2 & \rightarrow & \text{corresponds to point (-2,0)}
\\ x+8 = 0 & \longrightarrow & x=-8 & \rightarrow & \text{corresponds to point (-8,0)}
\end{array}\]
(To convince yourself, you can plug in each of those values of x one at a time and make sure the function equals zero for those values. ie
\[f(-8) = (-8-6)(-8+2)(-8+8)=(-14)(-6)(0)=0 \ \large \color{green}\checkmark\]

The last part of the question that asks you to find the y-intercept (by definition, where the line crosses the y axis; and also by definition where the x-axis equals zero) can be completed in a similar way to plugging in one of the values of the roots. All you have to do is substitute in x=0 for all x 's in the equation.

\[f(0) = (0-6)(0+2)(0+8)=(-6)(2)(8)= \text{y-intercept}\]

This is exactly like finding the y intercept of a normal straight line. For instance, if you had a straight line, with the equation of a line
\[y(x)=mx+b\]
you know that b is the y-intercept, because at x=0
\[ y(0)=m(0)+b=b=\text{y-intercept}\]

In your equation there are just a few more x terms.

Hope that this helped - holler if you have any questions! And good luck!! ^_^