## anonymous one year ago f(x)=2x2-x-3/x2-2x-3 graph and find the domain,range, and VA

1. anonymous

@mathmate

2. anonymous

no hold on :)

3. anonymous

|dw:1439337786064:dw|

4. mathmate

Do you mean $$\Large f(x)=\frac{2x^2−x−3}{x^2−2x−3}$$

5. anonymous

yes:)

6. mathmate

Do you know how to factorize quadratic expressions?

7. anonymous

yes!

8. anonymous

So I got for the first one, (x+1)(2x-3)

9. mathmate

So factor the numerator and denominator, as a start! ok?

10. anonymous

Got it:) So, I gottt (x+1)(2x-3)/(x+1)(x-3)

11. mathmate

Yep! Good job! So you have: $$\Large f(x)=\frac{(2x-3)(x+1)}{(x-3)(x+1)}$$ Do you notice anything that could help you?

12. anonymous

Canceling out the x+1

13. mathmate

Just like that, or with a condition?

14. anonymous

I dont know what you mean by that? So, I'm going w/ "just like that."

15. mathmate

Well, when you cancel factors in a division, you have to make sure that the factor will never become zero. So you can cancel ONLY if you specify x+1$$\ne$$0., in other words, x$$\ne-1$$.

16. mathmate

So what are you left with?

17. anonymous

2x-3/x-3 :)

18. mathmate

or $$\Large f(x)=\frac{(2x-3)}{(x-3)}, .... x\ne -1$$

19. mathmate

What else do you see?

20. anonymous

You can take out the x-3 and ur left with the 2

21. mathmate

You cannot simplify further, so you have to think of graphing what's left.

22. anonymous

Ok:)

23. mathmate

What graphing features do you see in the simplified function?

24. anonymous

ok hold on:)

25. anonymous

It goes in opposite directions

26. anonymous
27. mathmate

Well, have you learned about asymptotes?

28. anonymous

A horizontal asymptote for this may be 2/3

29. mathmate

Horizontal asymptote is correct, but not at 3/2. Horizontal asymptote can be obtained by taking limits of f(x) as x->inf and x->-inf.

30. mathmate

Do you see a vertical asymptote?

31. anonymous

no! Okay thats wrong! I mean 3,-3

32. mathmate

Almost! Vertical asymptotes are governed by the denominator becoming zero, so you need to find x for which (x-3)=0, or x=3. That's where the vertical asymptote is located.

33. mathmate

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34. mathmate

35. anonymous

hm...I give up on that one haha! So, I'm just going to guess and say 2/1

36. mathmate

You would divide the coefficients of x, so 2x/x=2, yes 2/1=2 is correct.

37. mathmate

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38. anonymous

:) Yay! Now what about the domain and range?

39. mathmate

We'll finish drawing the graph first. Do you know what it looks like?

40. anonymous

A point of intersection? Between the vertical and horizontal lines

41. mathmate

No this graph does not intersect itself, but it runs close to the two asymptotes that i drew.

42. mathmate

To know which quadrant the graph goes, you would evaluate f(x) at a point just to the left of x=3 (say 2.9) and a point just to the right of x=3, say 3.1 and find out where to plot the lines. So can you find f(2.9) and f(3.1)?

43. anonymous

=-0.2

44. mathmate

not really, f(2.9)=(2*2.9-3)/(2.9-3)=2.8/(-0.1)=-28 Your turn to do the same for f(3.1)...

45. anonymous

-32

46. mathmate

It's +32. The denominator is +0.1.

47. mathmate

So now we can complete the graph, remembering that $$x\ne-1$$ !

48. mathmate

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49. mathmate

Can you now do the domain and range?

50. anonymous

I'll try. So y>-1 and the domain could be all real numbers except 3

51. mathmate

How about -1 which we have shown an empty circle?

52. anonymous

Idk:(

53. mathmate

We said to cancel the common factor (x+1), we need to put the restriction $$x+1\ne 0$$ or $$x\ne -1$$, so =1 cannot show up in the domain, right?

54. anonymous

Right!

55. mathmate

*-1 So what is the domain?

56. anonymous

there would be one?

57. mathmate

Your proposed domain was almost right, except that you missed out the -1, so...

58. anonymous

Ok:) Thanks<3 I thought so, but I wasnt sure

59. mathmate

So what is it?

60. anonymous

all reals except 3,-1

61. mathmate

Good! Now proceed with the range, and you'd be done!

62. anonymous

Thank goodness lol!

63. mathmate

So what's the range?

64. anonymous

**Silence** Takes a wild guess and goes with...y>2/3

65. mathmate

Look at the graph for a start! There are two points to be excluded.

66. anonymous

2 and 3?

67. mathmate

We're looking at range, or the y-axis. So look for y-values that we cannot get from the graph!

68. anonymous

On the graph? ok so -1

69. mathmate

-1 is for x. What is the corresponding value for y????

70. mathmate

I.e. what is f(x) as x -> -1?

71. anonymous

i gotta go eat dinner! Thanks so much for all your help:). I'll be back in 30 minutes or so:)

72. mathmate

I will be gone by then. Someone else should be able to continue! Bon apetite!

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