Help with calculus!

- anonymous

Help with calculus!

- Stacey Warren - Expert brainly.com

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- anonymous

##### 1 Attachment

- freckles

just take the terms with highest degree from top and bottom

- freckles

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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## More answers

- anonymous

x^2

- freckles

right so your model for the first one is 2x/x^2 which can be reduced

- anonymous

Oh so when they ask for power function end behavior model it's asking for the highest power?

- anonymous

And it would simplify to \[\frac{ 2 }{ x }\] right?

- freckles

yes and I guess you could name it...
g(x)=2/x
and you can find the horizontal asymptote from the model

- freckles

\[\lim_{x \rightarrow \pm \infty } \frac{2}{x}=?\]

- freckles

here are more examples:
http://chargermath.wikispaces.com/file/view/2-2+End+Behavior+Models.pdf

- anonymous

Wouldn't the horizontal asymptote be 0 when graphed?

- freckles

yep
the horizontal asymptote would be y=0

- freckles

for the first one

- anonymous

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)?

- anonymous

Okay. So for number 2 from my examples g(x)=2 and the horizontal asymptote equals.....2?

- freckles

\[f(x)=\frac{4x^2-5x+6}{1} \\ g(x)=\frac{4x^2}{1}=4x^2\]

- anonymous

Oh okay that makes sense for the link problem 1.

- freckles

yes g(x)=2 would be the model

- freckles

and to find the horizontal aymsptote just evaluate:
\[y=\lim_{x \rightarrow \pm \infty}2\]

- anonymous

So it isn't 2?

- freckles

yep

- freckles

well y=2

- freckles

horizontal equations come in the form y=a number

- anonymous

So the actual answer would be y=2?

- freckles

just as the first one you asked about was y=0

- freckles

by the way the one in the link doesn't have a horizontal asymptote

- freckles

well the first one in that link I sent you

- anonymous

How did they get 1 as the final answer for problem 1 in the link?

- freckles

they were just checking the right and left conditions

- anonymous

Why?

- freckles

to see if it was a power function end behavior model

- anonymous

So that step proves that using the power method is correct?

- freckles

yes
as the pdf stated we need both
\[\lim_{x \rightarrow \pm \infty}\frac{f(x)}{g(x)}=1 \]

- freckles

this is to see f and g have the same kind of end behavior at both ends

- freckles

and if that holds then yes it is end behavior model

- anonymous

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

- anonymous

35.) g(x)=x^2; H.A.: y=0

- freckles

no ha

- anonymous

36.) g(x)=x; H.A.: DNE or undetermined?

- anonymous

Which part did I get wrong for 35?

- freckles

no ha
as in no horizontal aymptote

- anonymous

How do you find the horizontal asymptote numerically? I don't think I'm understanding how to find it using the graphing method.

- freckles

there is no polynomial that has a horizontal asymptote

- anonymous

Oh okay. Was I right for the others?

- freckles

|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

- freckles

just say dne

- freckles

I should add what kind of line I was talking about
horizontal line

- freckles

|dw:1441235738250:dw|
see the curve getting closer to that dark horizontal line I drew
that line being y=0

- freckles

|dw:1441235784454:dw|

- freckles

as we move out further and further from the origin in both left and right directions

- anonymous

Oh okay I got it. Thanks so much!

- freckles

np
so the only real correction you need to make
35 and 36 do not have horizontal asymptotes

- freckles

g(x)=x^2 and g(x)=x for 35 and 36 respectively look great

- anonymous

Okay

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