x-intercepts (bounce vs cut)

[sysD]

 

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[math]0.5((x-3)^3 - 3(x-3)^)[/math]

 

 

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THE LAST SAMURAI (not really)

Tonight, I ventured into the unknown. The shaman spoke, and the world faded; four arrows into Oblivion; a desolate landscape bereft of life and love. Suddenly, a polynomial appears!

Its teeth gnash. Half as short as a wild polynomial, this creature was clearly domesticated.

 

Undeniably though, malice glinted from its bloodshot eyes.

Steeling my will, I began to stare directly into the eyes of the polynomial.

Sweat pours down my face, the struggle of wills so devastatingly intense that surrounding air began to crackle with faint flickers of arcing energy.

The struggle is so brutal that it tears a hole in the fabric of spacetime, reavealing an alternate plane of reality consisting entirely of marble cheese.

This rift exists only for seconds - just long enough for indescribably complex extradimensional beings to notice this intrusion into their world. Whereupon, they siezed nearby litter and hurled it through the portal.

 

A direct consequence of this is that, a cardboard box flickers into existence.

 

The box is small; less than a square foot in diameter. Traffic raced inches to either side of it.

 

A kitten crawled out.

Traffic screeched - two tonnes of dead metal reared and bucked, its pilot attempting to avoid decimating this innocent creature. The SUV rolls, narrowly missing the kitten's outstretched paw... a mobile (pun intended) of doomed passengers.

In the innocent reflection of the eyes of a kitten, the SUV bursts into flame.

The driver screams.

The kitten mews.

 

The collar around its neck glints in the firelight.

Two local yokels retrieve the kitten.

"Hey lookie here, this lil fella's got an owner."

"Schrödinger...?"

"Like the sausage?"

"That's what she said..."

 

 

 

For those of you with short attention spans, I'm still involved in a life-or-death mental struggle with a vicious polynomial.

 

I vomit.

The strain will soon end me.

 

Wait! There - a root, wait, no, two!

 

Time slows.

For an instant, it stops completely.

 

 

My eyes widen, pupils following suit.

 

 

 

 

 

Within the eyes of this wretched creature, I have learned the truth.

 

From its gaping maw, a cry of derision.

 

 

 

 

 

My fists began to shake.

 

 

A snare thrown by the math gods;

 

this graph will both cut and bounce off of the of x-axis.

 

 

But Where?

 

 

 

(And how do I find out simply by plotting)

 

 

 

 

 

Next is the exciting conclusion of The Last Samurai (hrrrm). Find out if our hero survives this encounter - and also, the devestating fate of Schrödinger's kitten!

Edited April 11, 2011 by sysD

[mooeypoo]

sysD do you have an actual question? It's hard to tell within all this pseudospiritual text. Just ask.

 

 

Also, could you please mark all your text as size 1? It's hard to read.

[sysD]

Pseudospiritual? Its abstract math, bro.

 

Anyways, its all summed up in the bold:

[math]0.5((x-3)^3 - 3(x-3)^)[/math]

 

where's the bouncing intercept and where's the cutting intercept?

[Fuzzwood]

1) Where does this function equal 0? That is where your eq passes through the x axis?

 

2) Where does the differential eq equals 0? There will your maxima and minima be

 

3) Is it a maxima or minima? Differentiate a second time to see if you end up with a positive or negative equation.

[Cap'n Refsmmat]

What are "bouncing" and "cutting" intercepts? If I Google for that terminology it just leads me here. I don't know what you're asking for.

[sysD]

1) Where does this function equal 0? That is where your eq passes through the x axis?

 

2) Where does the differential eq equals 0? There will your maxima and minima be

 

3) Is it a maxima or minima? Differentiate a second time to see if you end up with a positive or negative equation.

 

Can't use calculus for this one =(

 

 

What are "bouncing" and "cutting" intercepts? If I Google for that terminology it just leads me here. I don't know what you're asking for.

 

Well, if you look at the graph of a function, terms with even exponents will "bounce" off of the axis. Terms with odd exponents will "cut" through the axis.

 

So, why does one term bounce and the other cut and where?

 

Oh, and...

http://www.wolframalpha.com/input/?i=y+%3D+.5%28%28x-3%29^3-3%28x-3%29^2%29

 

Wolfram states a y-intercept of -2

 

My calculations say it should be at -24.....

Edited April 11, 2011 by sysD

[mooeypoo]

Can't use calculus for this one =(

Why not? If you're looking for intercepts or if you're looking for how to graph this function, calculus is your method.

 

Well, if you look at the graph of a function, terms with even exponents will "bounce" off of the axis. Terms with odd exponents will "cut" through the axis.

This is why using proper definitions is crucial. We have no clue what you mean otherwise.

You seem to be looking for the end-behavior (set limits of the graph to infinity and negative infinity and see if the graph goes to a certain number or out to infinity) and the intercepts ("cutting" through the axes).

 

This is calculus.

 

So, why does one term bounce and the other cut and where?

I don't understand this question. Are you looking to graph this function?

 

Oh, and...

http://www.wolframal...3%28x-3%29^2%29

 

Wolfram states a y-intercept of -2

 

My calculations say it should be at -24.....

How did you calculate? Show the calculation so we can help or see if there's a mistake in it.

To get a y-intercept you set x=0. To get an x intercept you set y=0.

Show your calculation from this point and see if you still don't get what Wolfram Alpha gets.

 

 

~mooey

[sysD]

Why not? If you're looking for intercepts or if you're looking for how to graph this function, calculus is your method.

 

 

This is why using proper definitions is crucial. We have no clue what you mean otherwise.

You seem to be looking for the end-behavior (set limits of the graph to infinity and negative infinity and see if the graph goes to a certain number or out to infinity) and the intercepts ("cutting" through the axes).

 

This is calculus.

 

 

I don't understand this question. Are you looking to graph this function?

 

 

How did you calculate? Show the calculation so we can help or see if there's a mistake in it.

To get a y-intercept you set x=0. To get an x intercept you set y=0.

Show your calculation from this point and see if you still don't get what Wolfram Alpha gets.

 

 

~mooey

 

 

 

I can't use calculus because its a pre-calculus course. I'm supposed to be able to intuitively derive the general shape of the graph just by looking at the polynomial.

 

 

Finding y-intercept (0,y)

[math]f(0) = 0.5((0-3)^3 -3(x-3)^2)[/math]

 

[math]f(0) = 0.5((-3)^3 - 3(-3)^2)[/math]

[math]f(0) = 0.5(-27 - 3(9))[/math]

[math]f(0) = 0.5(-54)[/math]

[math]f(0) = -27[/math]

 

 

Edited April 12, 2011 by sysD

[Cap'n Refsmmat]

Wolfram does not state a y-intercept of -2. Note the label on the x-axis -- it's from 2 to 6, not 0 to 6. Your math is right.

[mooeypoo]

Finding y-intercept (0,y)

[math]f(0) = 0.5((0-3)^3 -3(x-3)^2)[/math]

 

[math]f(0) = 0.5((-3)^3 - 3(-3)^2)[/math]

[math]f(0) = 0.5(-27 - 3(9))[/math]

[math]f(0) = 0.5(-54)[/math]

[math]f(0) = -27[/math]

 

 

 

That's correct.

 

Your question regarding the "bouncing" and "cutting" intercepts are x-intercepts, though. To find these, you set y=0 this time, which is a bit more complicated because it's a cubic function. Still, there are a few things you can do (algebra, not calculus ) to make it easier and find the points of intercepts.

 

If you can try first, I can see how to help out more without giving you the answer outright. Don't let the cube power scare you.

[imatfaal]

sysD - whilst I don't disagree with Mooey or the Capt. above I thought I would add my five cents;

 

First off - think about what the 0.5 is doing to the function and especially when looking for roots.

And you are looking for roots - I interpret your bouncing as two roots that are the same (ie y=x² -4x +4 has two roots both at x=2), the line touches the axis but never cuts it (it stays either +ve or -ve) - and your cutting is when the curve intercepts the axis whilst moving from +ve to -ve or vice versa.

Once you realise what the 0.5 on the outside contributes you can simplify and that is the name of the game.

 

Secondly - the way the equation is given to you might ring some bells with some people that allow them to go straight to a solution , but if it is not then you need to grind it out. This means multiplying out brackets - putting like terms together to get standard polynomial form and then seeing if you can factor more normally. This is possible and doable for this function but boring.

 

Thirdly - as a very vague hint, once you have done it the boring way go to wolfram alpha and try changing every -3 for a -4 or a -5 and take note of how the roots change. Then perhaps try changing the moddle coefficient to -2 and -1. This should allow you to see a neat way of factorising that give you the answer directly

Edited April 13, 2011 by imatfaal

[Xittenn]

If you apply proper factoring and then answer the question, "which terms are even exponents and will "bounce" off of the axis and which terms are odd exponents and will "cut" through the axis?", this should be a no brainer. I mean Wolfram even gives you the proper form. And why is there any mention of y-intercept if the question clearly states this is with respect to the x-axis?

 

o.o

 

(imatfaal is my excuse for posting in this thread two days later)

[rattlesnake]

To find the x intercept you have to equate the Y=0 in the x,y equation as the usual coordinate of X intercept is (x,0)

and if you need to find Y intercept equate the x in the equation to zero

 

then find the maxima and minima value of the equation by differentiating the equation.

 

If you need more help you can surely log on to Tutorvista.com