MHB Questions regarding writing equations when given graphs

[mikea2424]

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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[skeeter]

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 $b=1$ ...

 

[mikea2424]

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))

 

[skeeter]

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

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