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Factoring pt2. rather than start a new topic i thought i'd ask my next question here. The expression is: 6x^2 + 11x - 10 How would I factor this?

d'oh. thanks, drrock, that really helped.

how would one get from: (2a+3)^2 - 8(2a+3) + 12 to: [math] [(2a + 3 ) -2] [ (2a+3) - 6] [/math] i expanded the first part and got: 4a^2 - 4a - 3

imatfall: thanks for the pointers So it seems things have gotten a bit off track. The corrected points are correct. The question never asked for the details of f(x), merely a sketch of its reciprocal. I found the reciprocal of "y" values to find the graph y=1/f(x) ------------------------------------------ Okay, here's what I have: The graph of f(x)=1/y Domain: xER, x cannot = -4, 0, 6 Range: yER, y cannot = 0 Therefore: Asymptotes @ (x=-4, x=0, x=6) & (y=0) The first quadrant of the graph (upper right) contains a line with points at (2, 1/2) and (4, 1/8). As "x" approaches 0 from the right, "y" approaches infinity. As "x" approaches infinity, "y" approaches 0. The second and fourth quadrants are empty. The third quadrant contains a downward facing parabola with the vertex at (-2, -1/2). As "x" approaches 0 from the left, "y" approaches negative infinity. As "x" approaches -4, "y" approaches negative infinity. __________________________________________________

I can't post the graph here due to copyright reasons... I can PM the document to you, though. Its Q.43 on the last page.

I'll try and be more specific on what exactly I'm having trouble with: The graph itself is shaped like a backwards "N" with points at (-4,0), (-2,-2), (0,0), (2,4), (4,8), (6,0) Turning points on the graph are at (-2,-2) and (4,8). Using the x-intercepts (y=0) for the unknown function, i formulated a basic equation. [math] f(x) = k(x+4)(x)(x-6) [/math] Solving for "k" using the point (4,8) yielded: [math] k = -1/8 [/math] The revised equation is: [math] f(x) = (-1/8)(x+4)(x)(x-6) [/math] Wolfram shows me this (which looks very similar to the graph given in the question: http://www.wolframalpha.com/input/?i=f%28x%29%3D%28-1%2F8%29%28x%2B4%29%28x%29%28x-6%29 However, its off in some key places. The places it is correct seem to be the points used to intuit the equation (4,8) and the intercepts. I'm going to try to incorporate some other points into this...

so how would i go about finding the function? or at least its recipricol?

Sorry. Seems the edit button has disappeared as well. The point is (-2,-2). not (-2,2)

I should have mentioned that the question gave me a rough graph. The points I copied down are correct. Here's a rough description of the graph: End Behaviour x --> (-) infinity, y --> (+) infinity y --> (+) infinity, y --> (-) infinity Turning Points local minima: (-2,-2) local maxima: (4,8)

could a mod change the topic title to polynomial recipricol?

Ey ppl. Here's the question: Given the the points on the graph y=f(x), find the equation of y=f(x) and y=1/f(x). Label asymptotes. Given Points: (-4,0), (-2, 2), (0,0), (2,4), (4,8), (6,0) Therefore: [math]f(x) = k(x)(x-6)(x+4)[/math] Using the local maxima (4,8), I began by solving the value of k. [math]f(4) = 8 = k(x)(x-6)(x+4)[/math] [math]8 = k(4)(-2)(8)[/math] [math]8 = -64k[/math] [math]k = 8/-64[/math] [math]k = -1/8[/math] Therefore: [math] k=(-1/8) [/math] The revised equation for f(x) is: [math] f(x) = -1/8(x+4)x(x-6) [/math] This is where I get stuck. This equation does not yield the ordered points that were given at the beginning of the question. I can't think of a vertical shift value © that would fix the equation. Any tips?

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]

Can't use calculus for this one =( 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.....

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?

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

Calculus is awesome and all, but I want to have an ironclad grasp on everything "pre-calculus."

Umm... sorry. That's kind of time consuming, especially on certain graphs.... Sounds promising, any leads? This is a good idea. I'll implement it in symmetrical graphs. I wish all of them were that easy

Hey there, I was just wondering if there is a simple way to find turning point (tan=0) coordinates without using calculus.

very nice.

Well, what I understood from that is: a and b are subsets of S. zero (the additive identity) is also included in S. the rest kind of went over my head. could you possibly break it down, maybe with an example??

haha i'll fax ya my visa =p

****I'm sorry, I figured out the answer to my own question. Is there a way to change the title of the thread?***** I have a related question: Title: Ternary Operators What are some examples of ternary operators in mathematics (not programming)?

http://news.blogs.cnn.com/2011/03/31/scientists-fight-flames-with-electric-wand/ "When the amplifier was turned on, the wand-like wire, serving as an electrode, generated an oscillating electric field that essentially pushed the flame off its fuel source. This extinguished it, said chemist and lead author Ludovico Cademartiri, a postdoctoral fellow who presented the findings in California on Sunday at a meeting of the American Chemical Society." This is so cool. Discuss?

I figured that the intense heat that close to the sun = more excited particles; therefore higher pressure. However, I forgot to take into account that Mercury has a virtually non-existent atmosphere, and pressure requires a medium to pressurize =p

Haha, I actually meant sticking a cod under the root system of a plant. As the cod decomposes and the nutrients become part of the soil, the plant growth greatly accelerates.