Help with calculus!

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Help with calculus!

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1 Attachment
just take the terms with highest degree from top and bottom
like for 2x+1 the term with highest degree is 2x for x^2-2x+1 the term with highest degree is?

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Other answers:

x^2
right so your model for the first one is 2x/x^2 which can be reduced
Oh so when they ask for power function end behavior model it's asking for the highest power?
And it would simplify to \[\frac{ 2 }{ x }\] right?
yes and I guess you could name it... g(x)=2/x and you can find the horizontal asymptote from the model
\[\lim_{x \rightarrow \pm \infty } \frac{2}{x}=?\]
here are more examples: http://chargermath.wikispaces.com/file/view/2-2+End+Behavior+Models.pdf
Wouldn't the horizontal asymptote be 0 when graphed?
yep the horizontal asymptote would be y=0
for the first one
In number 1 from the link you sent, how did they get g(x)=4x^2? Did they just use the 4x^2 from the f(x)?
Okay. So for number 2 from my examples g(x)=2 and the horizontal asymptote equals.....2?
\[f(x)=\frac{4x^2-5x+6}{1} \\ g(x)=\frac{4x^2}{1}=4x^2\]
Oh okay that makes sense for the link problem 1.
yes g(x)=2 would be the model
and to find the horizontal aymsptote just evaluate: \[y=\lim_{x \rightarrow \pm \infty}2\]
So it isn't 2?
yep
well y=2
horizontal equations come in the form y=a number
So the actual answer would be y=2?
just as the first one you asked about was y=0
by the way the one in the link doesn't have a horizontal asymptote
well the first one in that link I sent you
How did they get 1 as the final answer for problem 1 in the link?
they were just checking the right and left conditions
Why?
to see if it was a power function end behavior model
So that step proves that using the power method is correct?
yes as the pdf stated we need both \[\lim_{x \rightarrow \pm \infty}\frac{f(x)}{g(x)}=1 \]
this is to see f and g have the same kind of end behavior at both ends
and if that holds then yes it is end behavior model
Okay thanks! Can you check these 2 questions as well? Just want to make sure I'm getting this right. I'll say the answer I got and just check it.
1 Attachment
35.) g(x)=x^2; H.A.: y=0
no ha
36.) g(x)=x; H.A.: DNE or undetermined?
Which part did I get wrong for 35?
no ha as in no horizontal aymptote
How do you find the horizontal asymptote numerically? I don't think I'm understanding how to find it using the graphing method.
there is no polynomial that has a horizontal asymptote
Oh okay. Was I right for the others?
|dw:1441235528085:dw| but |dw:1441235564894:dw| do you see in the first graph the curve is approaching a line for really big values of x you can also see the graph is approaching a line for really big negative values of x that line being y=0 but nothing like that is happening in the second graph and yes you are right for the second question
just say dne
I should add what kind of line I was talking about horizontal line
|dw:1441235738250:dw| see the curve getting closer to that dark horizontal line I drew that line being y=0
|dw:1441235784454:dw|
as we move out further and further from the origin in both left and right directions
Oh okay I got it. Thanks so much!
np so the only real correction you need to make 35 and 36 do not have horizontal asymptotes
g(x)=x^2 and g(x)=x for 35 and 36 respectively look great
Okay

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