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anonymous

  • one year ago

f(x)=2x2-x-3/x2-2x-3 graph and find the domain,range, and VA

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  1. anonymous
    • one year ago
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    @mathmate

  2. anonymous
    • one year ago
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    no hold on :)

  3. anonymous
    • one year ago
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    |dw:1439337786064:dw|

  4. mathmate
    • one year ago
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    Do you mean \(\Large f(x)=\frac{2x^2−x−3}{x^2−2x−3}\)

  5. anonymous
    • one year ago
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    yes:)

  6. mathmate
    • one year ago
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    Do you know how to factorize quadratic expressions?

  7. anonymous
    • one year ago
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    yes!

  8. anonymous
    • one year ago
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    So I got for the first one, (x+1)(2x-3)

  9. mathmate
    • one year ago
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    So factor the numerator and denominator, as a start! ok?

  10. anonymous
    • one year ago
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    Got it:) So, I gottt (x+1)(2x-3)/(x+1)(x-3)

  11. mathmate
    • one year ago
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    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
    • one year ago
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    Canceling out the x+1

  13. mathmate
    • one year ago
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    Just like that, or with a condition?

  14. anonymous
    • one year ago
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    I dont know what you mean by that? So, I'm going w/ "just like that."

  15. mathmate
    • one year ago
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    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
    • one year ago
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    So what are you left with?

  17. anonymous
    • one year ago
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    2x-3/x-3 :)

  18. mathmate
    • one year ago
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    or \(\Large f(x)=\frac{(2x-3)}{(x-3)}, .... x\ne -1\)

  19. mathmate
    • one year ago
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    What else do you see?

  20. anonymous
    • one year ago
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    You can take out the x-3 and ur left with the 2

  21. mathmate
    • one year ago
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    You cannot simplify further, so you have to think of graphing what's left.

  22. anonymous
    • one year ago
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    Ok:)

  23. mathmate
    • one year ago
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    What graphing features do you see in the simplified function?

  24. anonymous
    • one year ago
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    ok hold on:)

  25. anonymous
    • one year ago
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    It goes in opposite directions

  26. mathmate
    • one year ago
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    Well, have you learned about asymptotes?

  27. anonymous
    • one year ago
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    A horizontal asymptote for this may be 2/3

  28. mathmate
    • one year ago
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    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.

  29. mathmate
    • one year ago
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    Do you see a vertical asymptote?

  30. anonymous
    • one year ago
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    no! Okay thats wrong! I mean 3,-3

  31. mathmate
    • one year ago
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    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.

  32. mathmate
    • one year ago
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    |dw:1439339170261:dw|

  33. mathmate
    • one year ago
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    What about the horizontal asymptote?

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

  35. mathmate
    • one year ago
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    You would divide the coefficients of x, so 2x/x=2, yes 2/1=2 is correct.

  36. mathmate
    • one year ago
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    |dw:1439339327350:dw|

  37. anonymous
    • one year ago
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    :) Yay! Now what about the domain and range?

  38. mathmate
    • one year ago
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    We'll finish drawing the graph first. Do you know what it looks like?

  39. anonymous
    • one year ago
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    A point of intersection? Between the vertical and horizontal lines

  40. mathmate
    • one year ago
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    No this graph does not intersect itself, but it runs close to the two asymptotes that i drew.

  41. mathmate
    • one year ago
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    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)?

  42. anonymous
    • one year ago
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    =-0.2

  43. mathmate
    • one year ago
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    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)...

  44. anonymous
    • one year ago
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    -32

  45. mathmate
    • one year ago
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    It's +32. The denominator is +0.1.

  46. mathmate
    • one year ago
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    So now we can complete the graph, remembering that \(x\ne-1\) !

  47. mathmate
    • one year ago
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    |dw:1439339824439:dw|

  48. mathmate
    • one year ago
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    Can you now do the domain and range?

  49. anonymous
    • one year ago
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    I'll try. So y>-1 and the domain could be all real numbers except 3

  50. mathmate
    • one year ago
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    How about -1 which we have shown an empty circle?

  51. anonymous
    • one year ago
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    Idk:(

  52. mathmate
    • one year ago
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    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?

  53. anonymous
    • one year ago
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    Right!

  54. mathmate
    • one year ago
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    *-1 So what is the domain?

  55. anonymous
    • one year ago
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    there would be one?

  56. mathmate
    • one year ago
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    Your proposed domain was almost right, except that you missed out the -1, so...

  57. anonymous
    • one year ago
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    Ok:) Thanks<3 I thought so, but I wasnt sure

  58. mathmate
    • one year ago
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    So what is it?

  59. anonymous
    • one year ago
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    all reals except 3,-1

  60. mathmate
    • one year ago
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    Good! Now proceed with the range, and you'd be done!

  61. anonymous
    • one year ago
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    Thank goodness lol!

  62. mathmate
    • one year ago
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    So what's the range?

  63. anonymous
    • one year ago
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    **Silence** Takes a wild guess and goes with...y>2/3

  64. mathmate
    • one year ago
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    Look at the graph for a start! There are two points to be excluded.

  65. anonymous
    • one year ago
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    2 and 3?

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

  67. anonymous
    • one year ago
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    On the graph? ok so -1

  68. mathmate
    • one year ago
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    -1 is for x. What is the corresponding value for y????

  69. mathmate
    • one year ago
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    I.e. what is f(x) as x -> -1?

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

  71. mathmate
    • one year ago
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    I will be gone by then. Someone else should be able to continue! Bon apetite!

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