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Mathematics 25 Online
OpenStudy (anonymous):

Medal and I'll fan you! Using those points as zeros, construct the polynomial function, f(x) that will be the path of your roller coaster. Show all of your work. That's the question and i made these my points : 1st point: (-11,8) 2nd point: (-5,-4) 3rd point : (7,7)

Directrix (directrix):

How did you generate these points, and if they are zeros of the polynomial, then why is the y-coordinate for each not zero? Do you anticipate that the polynomial function will be a quadratic?

OpenStudy (anonymous):

It just asked to pick 3 points that didn't have the buildings on them.

Directrix (directrix):

I don't understand this: Using those points as zeros, ... The points listed are not zeros. Maybe you could post the diagram.

OpenStudy (anonymous):

okay one second :)

OpenStudy (anonymous):

OpenStudy (anonymous):

@Directrix

Directrix (directrix):

Where are the instructions you were given about choosing 3 points that are zeros of the polynomial? These are not zeros. 1st point: (-11,8) 2nd point: (-5,-4) 3rd point : (7,7) It is correct that they do not have buildings on them.

Directrix (directrix):

What about three points such as: (11,0) (5,0) (-8,0)

Directrix (directrix):

Is this the question: As one of the new rollercoaster engineers, you have been tasked with developing a rollercoaster that will intertwine with existing Oakville Lake Amusement Park structures. For one of the more thrilling sections, the rollercoaster 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 rollercoaster crosses the x–axis. Precise measurements and attention to detail are very important. 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.) 1.The rollercoaster will run from right to left, so work in that order when selecting points. •1st point:______ •2nd point:______ •3rd point:______

OpenStudy (anonymous):

YES!

Directrix (directrix):

@Hannah28 Read through this (same problem plus solution) and see what you think. http://openstudy.com/study#/updates/526a9861e4b08ed552540a02 I'll read it, too.

OpenStudy (anonymous):

I read through that and I'm not 100% sure what to do still but I think I might have an idea...

Directrix (directrix):

Definitely, change those points so that they are zeros of the polynomial.

OpenStudy (anonymous):

so choose points on the x line?

Directrix (directrix):

The rollercoaster will run from right to left So the points should go in descending order.

OpenStudy (anonymous):

7, 4, -12?

Directrix (directrix):

Did you see this: the rollercoaster will dive down in-between buildings, plummet underground, pop back up, and coast over a hill before shooting back underground.

OpenStudy (anonymous):

Oh my gosh :( I didn't even see that!

OpenStudy (anonymous):

they used 5, -4, -12 so is there only one answer or could I use different points?

Directrix (directrix):

I think there will be different answers based on the zeros you choose. Use different points. Then, we will crank out the polynomial, draw its graph, and see if it goes under anything. These points you had at the start: 7, -5, -11 --> Do they run into any of the buildings?

Directrix (directrix):

-11 does

OpenStudy (anonymous):

-11 does

OpenStudy (anonymous):

so could I change that to -12?

Directrix (directrix):

Now, I'm thinking that the points have to be chosen to make the RC do this: the rollercoaster will dive down in-between buildings, plummet underground, pop back up, and coast over a hill before shooting back underground.

OpenStudy (anonymous):

will these points do that?

Directrix (directrix):

It may be doing this. What do you think?

OpenStudy (anonymous):

Is that what I want?

Directrix (directrix):

I don't know. I just went by this: the rollercoaster will dive down in-between buildings, plummet underground, pop back up, and coast over a hill before shooting back underground.

Directrix (directrix):

Now, what are the points that would go with that diagram?

Directrix (directrix):

(5,0); (-4,)); (-13,0)

OpenStudy (anonymous):

5 -4 -13?

OpenStudy (anonymous):

Yup!:)

Directrix (directrix):

Let's try: y = (x -5)*(x+4)*(x+13) Oh,no. Look at the graph here: http://www.wolframalpha.com/input /?i=y+%3D+%28x+-5%29*%28x%2B4%29*%28x%2B13%29 It is not doing this: shooting back underground.

OpenStudy (anonymous):

how is that not right? It looked like we want it to?

Directrix (directrix):

I was going by this: The rollercoaster will run from right to left. Now, I cannot find where I read it in the original problem.

OpenStudy (anonymous):

it is coming down from right though

Directrix (directrix):

But, the RC does not pop back up at x = -13

OpenStudy (anonymous):

is it supposed to?

OpenStudy (anonymous):

it looks like the diagram you made?

Directrix (directrix):

I think I have forgotten how to read: See if the RC graph is doing this: the rollercoaster will dive down in-between buildings, plummet underground, pop back up, and coast over a hill before shooting back underground.

OpenStudy (anonymous):

that's what its doing :)

Directrix (directrix):

We should check to see if the RC crashes into any of the three buildings.

OpenStudy (anonymous):

oh that's smart

OpenStudy (anonymous):

How do I know if it crashes or not?

Directrix (directrix):

You could evaluate y = (x-5)(x+4)(x+3) for x = 1.5, x = 9.5, and x = -10 to determine if the y values for each of those is higher than the buildings. Or, we could put a factor of 10 in front of the polynomial to make it taller. Let's see what that looks like.

OpenStudy (anonymous):

Those look like they work to me :)

Directrix (directrix):

The multiplication by 10 may take the FW into outer space. I don't know.

OpenStudy (anonymous):

Oh :( I'm really confused!

Directrix (directrix):

We can go with the original: y = (x -5)*(x+4)*(x+13) which as a polynomial is: P(x) = x³ + 12 x² - 33x - 260

OpenStudy (anonymous):

how did you get that equation?

Directrix (directrix):

www.wolframalpha.com is your friend. Click on this link: http://www.wolframalpha.com/input/?i=y+%3D+%28x+-5%29*%28x%2B4%29*%28x%2B13%29

Directrix (directrix):

You can multiply those binomials by hand to check.

OpenStudy (anonymous):

Im lost...

Directrix (directrix):

You asked this: how did you get that equation? I multiplied this y = (x -5)*(x+4)*(x+13) out to be in polynomial form. If 5 is a zero of the polynomial, then (x-5) is a factor of the polynomial. If -4 is a zero of the polynomial, the (x- (-4)) = (x + 4) is a factor of the polynomial. If -13 is a zero of the polynomial, then (x - (-13)) = (x+13) is a factor of the polynomial. So, the entire polynomial is the product of those three factors: (x -5)*(x+4)*(x+13). I used an online calculator to multiply all that. You can check that multiplication if you like by multiplying it out with paper and pen/pencil.

Directrix (directrix):

@Hannah28 Does this ^^^ make sense?

OpenStudy (anonymous):

I left for the beach sorry :( Thank you @Directrix

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