Question

Help with calculus!

Answer 1

just take the terms with highest degree from top and bottom

Answer 2

like for 2x+1
the term with highest degree is 2x
for x^2-2x+1 the term with highest degree is?

Answer 3

x^2

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

Wouldn't the horizontal asymptote be 0 when graphed?

Answer 11

yep
the horizontal asymptote would be y=0

Answer 12

for the first one

Answer 13

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)?

Answer 14

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

Answer 15

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

Answer 16

Oh okay that makes sense for the link problem 1.

Answer 17

yes g(x)=2 would be the model

Answer 18

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

Answer 19

So it isn't 2?

Answer 20

yep

Answer 21

well y=2

Answer 22

horizontal equations come in the form y=a number

Answer 23

So the actual answer would be y=2?

Answer 24

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

Answer 25

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

Answer 26

well the first one in that link I sent you

Answer 27

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

Answer 28

they were just checking the right and left conditions

Answer 29

Why?

Answer 30

to see if it was a power function end behavior model

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

and if that holds then yes it is end behavior model

Answer 35

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.

Answer 36

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

Answer 37

no ha

Answer 38

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

Answer 39

Which part did I get wrong for 35?

Answer 40

no ha
as in no horizontal aymptote

Answer 41

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

Answer 42

there is no polynomial that has a horizontal asymptote

Answer 43

Oh okay. Was I right for the others?

Answer 44

but

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