Questions regarding writing equations when given graphs

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Discussion Overview

The discussion revolves around writing equations for rational functions based on provided graphs. Participants are attempting to derive the equations from the characteristics of the graphs, including zeros, asymptotes, and intercepts. The scope includes mathematical reasoning and problem-solving related to pre-calculus concepts.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Technical explanation

Main Points Raised

  • One participant identifies a zero at $x=1$, a y-intercept at $(0,-1)$, and vertical asymptotes at $x=\pm2$, proposing a function form involving constants $a$ and $b$.
  • Another participant suggests that the zero at $x=1$ has even multiplicity, indicating the numerator should be $(x-1)^2$, and notes vertical asymptotes at $x=\pm3$.
  • There is uncertainty regarding the horizontal asymptote, with one participant stating that the degree of the numerator must be less than that of the denominator.
  • One participant proposes a function form $f(x) = \dfrac{a(x-1)^2}{b(x+3)(x-3)^2}$, indicating that the factor $(x-3)$ in the denominator is squared based on the function's behavior near the vertical asymptote.
  • Participants express difficulty in determining the correct values for constants $a$ and $b$, with one suggesting $a=1$ and calculating $b$ based on a specific point on the graph.

Areas of Agreement / Disagreement

Participants generally agree on the presence of certain features in the graphs, such as zeros and asymptotes. However, there is disagreement and uncertainty regarding the exact forms of the equations and the behavior of the functions, particularly concerning the horizontal asymptote and the multiplicity of factors.

Contextual Notes

Participants note the lack of additional points on the graphs to definitively determine the constants $a$ and $b$, which complicates the formulation of the equations.

mikea2424
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Hello,

I'm studying for a test this Friday and I have two problems that have completely stumped me. I'm not sure how to solve either. Any help would be greatly appreciated.

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Mathematics news on Phys.org
http://mathhelpboards.com/attachments/pre-calculus-21/7234d1503462274-questions-regarding-writing-equations-when-given-graphs-1-jpg

you should know these are both rational functions ... I leave the second graph for you to try.

for the first graph ...
(1) single zero at $x=1$
(2) y-intercept at $(0,-1)$
(3) horizontal asymptote appears to be $y=0$
(4) vertical asymptotes at $x=\pm2$, with the graph $\to +\infty$ for asymptote at $x=2$

$f(x)=\dfrac{a(x-1)}{b(x+2)(x-2)^2}$, where $a$ and $b$ are constants

$f(0)=-1=-\dfrac{a}{8b} \implies a=8b$. To fix $a$ and $b$ requires another point on the graph. I'm not going to try and eyeball another point, so to keep it simple, let $b=1$ ...
 
Thank you very much for replying to my original post.
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I know that there is a zero at x=1. I know this zero touches the x-axis meaning that it is an even multiplicity. So the top of the fraction should look like (x-1)^2. The vertical asymptotes are x= \pm3. So the bottom of the fraction looks like (x+3)(x-3). This is where I get confused..visibly I can't tell what the horizontal asymptote is. I also am unsure which of the bottom factors I am to square, if any at all. From my notes I can gather that if the parabola goes downward as well as the curves then I am to square one of the factors on the bottom. I have tried squaring each, one at a time and input my equation into Desmos with no luck. As of right now all I know for sure is:

a((x-1)^2)/((x+3)(x-3))
 

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The fact there is a horizontal asymptote tells you (numerator degree) < (denominator degree)

As of right now all I know for sure is:

a((x-1)^2)/((x+3)(x-3))

note the function's behavior near the vertical asymptote $x = 3$ indicates the value $(x-3)$ in the denominator is squared ,,,

$f(x) = \dfrac{a(x-1)^2}{b(x+3)(x-3)^2}$

$f(0) = -0.4 \implies \dfrac{a}{27b} = -\dfrac{2}{5} \implies b = -\dfrac{5a}{54}$

once again, not having a second point on the curve other than an x or y intercept ...

let $a = 1 \implies b = -\dfrac{5}{54}$ ...

$f(x) = -\dfrac{54(x-1)^2}{5(x+3)(x-3)^2}$
 

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