sysD Posted August 25, 2011 Share Posted August 25, 2011 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? Link to comment Share on other sites More sharing options...

sysD Posted August 25, 2011 Author Share Posted August 25, 2011 could a mod change the topic title to polynomial recipricol? Link to comment Share on other sites More sharing options...

CraXshot Posted August 25, 2011 Share Posted August 25, 2011 Okay i'm not shure. Are you shure it's polynomial? Remember those x-axis points may be qwadratic so that the graph only touches them.In other words (0;0) might be a turning point wich means it should be f(x)=x^{2}(...) or any one of the other points. if you can confirm that it's a polynomial we know that it can only have 2 turning points. One somewhere where -2<x<2 and 2<x<6, and if you're shure that (4,8), is the local maximum(how do you know that?) we only need the one between -2<x<2. I've tried it with (0;0) as a turningpoint, you can too just too check me but it didn't work out. f(x) still didnt give all the points so therefore there must be another x-axis intercept between -2 and 2. Sorry if this is confusing but i'm just thinking outloud and hoping to give you some ideas but so far it looks to me like you're information isn't sufficient. Did they say something more about those co-ordinates(turning points or like you said local maximum etc.) Link to comment Share on other sites More sharing options...

DrRocket Posted August 25, 2011 Share Posted August 25, 2011 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? In general a polynomial is not determined solely by its real zeros, even up to a multiple. So while [math] (x+4)x(x-6) [/math] is a factor of the specified polynomial, it is not the whole thing. In general given n points there will be a polynomial of degree n-1 passing through those points. There will also be polynomials of higher degree, so the points that you have listed do not define a unique function. Link to comment Share on other sites More sharing options...

Fuzzwood Posted August 25, 2011 Share Posted August 25, 2011 The way k(x) is written, it suggests another function. Link to comment Share on other sites More sharing options...

sysD Posted August 25, 2011 Author Share Posted August 25, 2011 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) Link to comment Share on other sites More sharing options...

CraXshot Posted August 26, 2011 Share Posted August 26, 2011 Quote: Turning Points local minima: (-2,-2) local maxima: (4,8) 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) How can (-2,-2) and (-2, 2) both be points on the graph, then you don't have a fuction Link to comment Share on other sites More sharing options...

imatfaal Posted August 26, 2011 Share Posted August 26, 2011 (edited) Craxshot is correct - but I assume it is a typo. The y value at x=-2 is (I guess) -2 . This potentially allows only 2 turning points at the coordinates given - if the point is (-2,2) then three turning points are a minimum and equation cannot be a cubic - and whilst Dr R is correct, I am willing to bet that it is otherwise question is endless. bearing in mind that there is at least one typo - could the OP check all the points? hey, if possible put up a scan of the question Further more - and I stand to be corrected for faulty memory/intuition - are not cubics point symmetrical around the point of inflection (which must be x=1); and these points clearly are not symmetrical around point at x=1. So either Dr R is correct and it is just a portion of a higher order polynomial (of which you have found a lower order polynomial factor) - or you have more than one typo in the OP Edited August 26, 2011 by imatfaal Link to comment Share on other sites More sharing options...

sysD Posted August 26, 2011 Author Share Posted August 26, 2011 Sorry. Seems the edit button has disappeared as well. The point is (-2,-2). not (-2,2) Link to comment Share on other sites More sharing options...

sysD Posted August 26, 2011 Author Share Posted August 26, 2011 so how would i go about finding the function? or at least its recipricol? Link to comment Share on other sites More sharing options...

DrRocket Posted August 26, 2011 Share Posted August 26, 2011 so how would i go about finding the function? or at least its recipricol? n points on a graph give you n equations for the n coefficients of a polynomial of degree n-1. Link to comment Share on other sites More sharing options...

sysD Posted August 27, 2011 Author Share Posted August 27, 2011 (edited) 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... Edited August 27, 2011 by sysD Link to comment Share on other sites More sharing options...

imatfaal Posted August 30, 2011 Share Posted August 30, 2011 sysD - I think you have more points wrong (either typos or in the question as given). Either as DocRock explained you have a question asking for high order equations, or if you are meant to be looking for a cubic (ie x to power three) then there is something wrong. On your graph - of course the x axis intercepts are correct, they are the roots that you based your equation on - they could not be wrong. The turning points are wrong! [math] f(x) = (-1/8)(x+4)(x)(x-6) [/math] [math] y=\frac{-x^3}{8}+\frac{x^2}{4}+3x[/math] [math] \frac{dy}{dx} = \frac{-3x^3}{8}+\frac{x}{2}+3[/math] local minima and maxima will be when derivative equals zero - using quadratic solution when x equals [math] \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/math] [math] \frac{-1/2 \pm \sqrt{(1/2)^2-4(-3/8)(3)}}{2(-3/8)}[/math] simplify to (by multiplying through by -8/2) [math] \frac{2 \pm -4(\sqrt{(2/8)+(36/8)})}{3}[/math] sort out sqrt [math] \frac{2 \pm -4\sqrt{19/4}}{3}[/math] take a factor of two into sqrt [math] \frac{2 \pm 2\sqrt{19}}{3}[/math] and separate for ease of reading [math] \frac{2}{3} \pm \frac{2\sqrt{19}}{3}[/math] Now those turning points do not meet up with yours - ie (4,8) is no longer a turning point and neither is (-2,-2) Link to comment Share on other sites More sharing options...

sysD Posted August 30, 2011 Author Share Posted August 30, 2011 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. Link to comment Share on other sites More sharing options...

sysD Posted September 3, 2011 Author Share Posted September 3, 2011 (edited) 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. __________________________________________________ Edited September 3, 2011 by sysD Link to comment Share on other sites More sharing options...

sysD Posted September 12, 2011 Author Share Posted September 12, 2011 I've uploaded a rough sketch of the graph. http://www.megaupload.com/?d=4OMOS0T9 Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now