## anonymous one year ago Help with calculus!

1. anonymous

2. freckles

just take the terms with highest degree from top and bottom

3. freckles

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

4. anonymous

x^2

5. freckles

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

6. anonymous

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

7. anonymous

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

8. freckles

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

9. freckles

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

10. freckles

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

11. anonymous

Wouldn't the horizontal asymptote be 0 when graphed?

12. freckles

yep the horizontal asymptote would be y=0

13. freckles

for the first one

14. 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)?

15. anonymous

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

16. freckles

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

17. anonymous

Oh okay that makes sense for the link problem 1.

18. freckles

yes g(x)=2 would be the model

19. freckles

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

20. anonymous

So it isn't 2?

21. freckles

yep

22. freckles

well y=2

23. freckles

horizontal equations come in the form y=a number

24. anonymous

So the actual answer would be y=2?

25. freckles

26. freckles

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

27. freckles

well the first one in that link I sent you

28. anonymous

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

29. freckles

they were just checking the right and left conditions

30. anonymous

Why?

31. freckles

to see if it was a power function end behavior model

32. anonymous

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

33. freckles

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

34. freckles

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

35. freckles

and if that holds then yes it is end behavior model

36. 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.

37. anonymous

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

38. freckles

no ha

39. anonymous

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

40. anonymous

Which part did I get wrong for 35?

41. freckles

no ha as in no horizontal aymptote

42. anonymous

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

43. freckles

there is no polynomial that has a horizontal asymptote

44. anonymous

Oh okay. Was I right for the others?

45. 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

46. freckles

just say dne

47. freckles

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

48. freckles

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

49. freckles

|dw:1441235784454:dw|

50. freckles

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

51. anonymous

Oh okay I got it. Thanks so much!

52. freckles

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

53. freckles

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

54. anonymous

Okay