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anonymous

  • one year ago

Graph f(x)=x^2-3x-10/x-2. Find the domain, range, VA, HA/SA

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  1. misty1212
    • one year ago
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    HI!!

  2. anonymous
    • one year ago
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    Hey:)

  3. misty1212
    • one year ago
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    is it \[\frac{x^2-3x-10}{x-2}\]

  4. anonymous
    • one year ago
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    Yes

  5. misty1212
    • one year ago
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    ok lets do them one at a time

  6. misty1212
    • one year ago
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    domain is easy enough, set the denominator equal to zero and solve then say "all real numbers except that one"

  7. misty1212
    • one year ago
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    what do you get for that? (you can do it in your head)

  8. anonymous
    • one year ago
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    2

  9. misty1212
    • one year ago
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    yeah so all numbers except 2

  10. misty1212
    • one year ago
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    now before we continue, lets make sure this is not a trick question can you factor the numerator and cancel?

  11. anonymous
    • one year ago
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    Let me see...

  12. misty1212
    • one year ago
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    hmmi think it does factor, but not cancel !

  13. anonymous
    • one year ago
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    (x-5)(x+2)/x-2

  14. misty1212
    • one year ago
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    yeah but nothing cancels, so lets continue but you should always check that first

  15. misty1212
    • one year ago
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    once we have the domain, we also have the vertical asymptote too

  16. misty1212
    • one year ago
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    clear or no?

  17. anonymous
    • one year ago
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    So it is also 2

  18. anonymous
    • one year ago
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    yes!

  19. misty1212
    • one year ago
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    careful here

  20. misty1212
    • one year ago
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    the vertical asymptote is a vertical line, not a number

  21. misty1212
    • one year ago
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    so you want to return and answer of \(x=2\) the vertical line, not just the number 2

  22. misty1212
    • one year ago
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    clear?

  23. anonymous
    • one year ago
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    Oh! Yes! Clear

  24. misty1212
    • one year ago
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    now for the horizontal asympote the degree of the numerator is ?

  25. anonymous
    • one year ago
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    2

  26. misty1212
    • one year ago
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    and the denominator?

  27. anonymous
    • one year ago
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    there is no horizontal then since it's greater than the denom.

  28. misty1212
    • one year ago
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    bingo

  29. misty1212
    • one year ago
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    but the degree of the numerator is one more than the degree of the denominator that means there will be a slant asymptote do you know how to find it?

  30. anonymous
    • one year ago
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    I never learned about a slant asymptote so I dont know :(

  31. misty1212
    • one year ago
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    ok it is easy to say what to do , but not easy for me to write in one word "divide"

  32. misty1212
    • one year ago
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    when you divide a polynomial of degree 2 by a polynomial of degree 1 you get a polynomial of degree 1 ( a line) and a remainder ignore the remainder, the line is the slant asymptote

  33. misty1212
    • one year ago
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    you can use either long division or (much easier) synthetic division do you know how to do that?

  34. anonymous
    • one year ago
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    Yes :)

  35. misty1212
    • one year ago
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    whew like i said it is very hard for me to write division here what do you get?

  36. misty1212
    • one year ago
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    take your time

  37. anonymous
    • one year ago
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    I did it in a rush so I got x^2-1x-6.

  38. misty1212
    • one year ago
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    hmm no you should have a polynomial of degree 1, not a quadratic

  39. misty1212
    • one year ago
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    1 -3 -10 2 ______________ 1

  40. misty1212
    • one year ago
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    that is the iniital set up then 1 -3 -10 2 2 ______________ 1 -1

  41. misty1212
    • one year ago
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    the next line is unimportant since we only need the line, not the remainder the line has the coefficients from the final row so \(x-1\) slant asymptote is \(y=x-1\)

  42. anonymous
    • one year ago
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    Oh! Now I know what I did wrong! When did the fak divison, i fogot to add the "0^2" Lol but yes, I know understand

  43. misty1212
    • one year ago
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    ok then we are done!

  44. misty1212
    • one year ago
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    oh except for the range

  45. anonymous
    • one year ago
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    I'm so bad at finding that.

  46. misty1212
    • one year ago
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    in this case it is real easy, since it has a vertical asympote and no horizontal asymptote that means it goes from \(-\infty\) to \(\infty\)

  47. misty1212
    • one year ago
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    you can write your answer as an interval \((-\infty, \infty)\) or just say "all real numbers"

  48. misty1212
    • one year ago
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    it is harder for some other ones, but not for this one

  49. anonymous
    • one year ago
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    Yay! Thanks so much:)))

  50. misty1212
    • one year ago
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    \[\color\magenta\heartsuit\]

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