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## anonymous 2 years ago PLEASE HELP :( Please help: sketch the graph of a function. label the intercepts, relative extrema, points of inflection, and asymptotes! then state the domain of the function! y= (5-3x)/(x-2)

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1. anonymous

so i found the vertical asymptote: x=2 and horizontal asymptote: y=3 and the intercepts i found that it is (0, -2.5) and (1.67,0) so the domain is: (-inf,2)(2, inf) now I'm stuck on how to do the relative extrema and the points of inflections. :( please help.

2. anonymous

i tried to also find the derivative and i got 1/ (x-2)^2

3. anonymous

@zepdrix please help

4. zepdrix

So did you apply the Quotient Rule? One sec, I'll check it :O

5. anonymous

yeah i used the quotient rule

6. zepdrix

Mmm yah I think your first answer was correct: 1/ (x-2)^2

7. anonymous

yeah the first answer was correct.

8. anonymous

i mean I'm trying to find the relative extrema but i forgot how to. i know i have to find the derivative first. then I'm not sure where to go . :(

9. zepdrix

So we have our first derivative:$\Large y'\quad=\quad \frac{1}{(x-2)^2}$ To find extrema/critical points, we set the first derivative equal to zero and solve for x.

10. anonymous

right but wouldn't it be false?

11. anonymous

1=0? doesn't work.

12. zepdrix

Oh there is another little tid bit that I often forget. critical points also exist when the first derivative is undefined. Do you see any points where that might be true?

13. anonymous

by undefined what do you mean?

14. zepdrix

Numbers that would cause a problem :D Like dividing by zero.

15. zepdrix

Any x values that would cause a problem for our derivative?

16. anonymous

like this one?

17. anonymous

oh ya 2!

18. zepdrix

Yes! Good. x=2 is something we might consider to be a critical point. Before we can say that for certain though, is x=2 in the domain of our function?

19. zepdrix

You stated the domain earlier, so go back and look at it! :)

20. anonymous

no.

21. anonymous

x cannot equal to 2.

22. zepdrix

Good good. Since it's not in the domain of our function, it can't be a critical point.

23. anonymous

all real numbers except 2.

24. zepdrix

So there are no critical points. Should we look for inflection points next? :o

25. anonymous

so theres no relative extrema?

26. anonymous

yes last but not least, inflection points. :)

27. zepdrix

So we'll need to find the second derivative for that. Have you tried that step yet?

28. anonymous

no but let me try real quick

29. anonymous

so inflection points is always SECOND DERIVATIVE right?

30. zepdrix

y=0 gives us roots of the function. y'=0 gives us critical points of the function. y''=0 gives us inflection points of the function. yes.

31. anonymous

i got (-3x^3+17x^2-32x+20)/(x^2-4x+4)

32. anonymous

@zepdrix

33. zepdrix

Dude dude dude -_- what..? Don't expand out squares, especially if they're in the denominator.

34. zepdrix

Did you apply the quotient rule? We don't want to do that.

35. anonymous

yeah i did:/

36. anonymous

i used the rule book.

37. zepdrix

$\Large y'\quad=\quad \frac{1}{(x-2)^2}\quad=\quad (x-2)^{-2}$Power rule from there :O

38. anonymous

ohhhhh yes

39. anonymous

crap.

40. zepdrix

XD

41. anonymous

-2/(x-2)^3

42. anonymous

ok :D

43. zepdrix

Ok cool. Setting our second derivative equal to zero:$\Large 0\quad=\quad \frac{-2}{(x-2)^3}$

44. zepdrix

Hmm I think inflection points work the same way as critical points ~ they exist where the second derivative is undefined.. mmmmm

45. anonymous

huh?

46. anonymous

don't we plug in severals x's between the intervals or something?

47. anonymous

so threres no inflection point?

48. zepdrix

between what intervals though? We would need to find inflection points to determine which intervals to check. I guess we just check where the function has a discontinuity since we have no inflection points. So ya, check on the left side of 2 and also on the right. We should get different concavity on each side for this problem.

49. zepdrix

Oh the problem didn't ask for concavity did it? -_- oh.. hmm

50. anonymous

no it didn't...

51. anonymous

so help me with inflection point, do we have to include other x's between the intervals?im not sure

52. zepdrix

include other x's... whu? +_+

53. anonymous

so this is the answer: it doesn't say which is for which but i got the idea of what is what. my question is how'd they get the other curve on the right side

54. zepdrix

So they labeled the intercepts (as you found), the asymptotes x=2, y=-3, and nothing else... So I think we did everything correctly, seeing as how we found no extrema for this one and they didn't label any on their graph.. :\

55. zepdrix

Oh the other curve, umm

56. zepdrix

bah i dunno >:O my head hurts.. i need a maf break!

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