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anonymous

  • one year ago

Graph 3/x^2-4. Then find the domain/range/VA

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  1. Astrophysics
    • one year ago
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    \[f(x) = \frac{ 3 }{ x^2 }-4?\] is this your equation, think of the parent function, which in this case is \[\frac{ 1 }{ x^2 }\] it looks like |dw:1439265308847:dw| you can get this by simply making a table of values which you can do yourself. Once you have your parent function now we can do our transformations. \[y = \frac{ 3 }{ x^2 }\] where the 3 should give you a scaling factor then afterwards add the -4, \[y = \frac{ 3 }{ x^2 }-4\] where the -4 should give you a vertical translation of -4 units. If you're not comfortable at seeing it directly you can always make a table of values for each step, good luck!

  2. anonymous
    • one year ago
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    No:( It is |dw:1439336098501:dw|

  3. anonymous
    • one year ago
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    I'm sorry lol:)

  4. Astrophysics
    • one year ago
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    That means you're graph will look like this then |dw:1439336233138:dw|

  5. Astrophysics
    • one year ago
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    Actually here, use this site to graph it out, but make sure you know how the transformations work https://www.desmos.com/calculator

  6. anonymous
    • one year ago
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    Thanks:)! What about the domain, range

  7. Astrophysics
    • one year ago
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    For domain solve \[x^2 - 4 \neq 0\] solve for x

  8. anonymous
    • one year ago
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    um -2 or 2?

  9. Astrophysics
    • one year ago
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    Yup, exactly

  10. Astrophysics
    • one year ago
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    Notice that domain occupies the x values, and range is the y values so you have y>0 and \[y \le -3/4\]

  11. anonymous
    • one year ago
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    ok:) And what about the VA? I have no idea what that means haha

  12. Astrophysics
    • one year ago
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    vertical asymptotes

  13. Astrophysics
    • one year ago
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    I'm assuming you don't know what limits are, so you see where it's not defined

  14. Astrophysics
    • one year ago
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    So VA is at x = 2 and x = -2

  15. anonymous
    • one year ago
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    And last one, I promise :) It says find the HA/SA???

  16. Astrophysics
    • one year ago
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    I don't know what SA is but HA is horizontal asymptote

  17. anonymous
    • one year ago
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    Okay so would that be -3/2?

  18. Astrophysics
    • one year ago
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    Not quite, since the denominator has a higher degree than the numerator it's 0

  19. Astrophysics
    • one year ago
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    That's just a shortcut though

  20. Astrophysics
    • one year ago
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    You should actually look up what exactly this stuff means haha

  21. anonymous
    • one year ago
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    lol! I probably should :) and I will

  22. anonymous
    • one year ago
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    so just checking, so far we have: VA=-2,2 D-all real numbers r-y<=-3/4

  23. Astrophysics
    • one year ago
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    domain is not all real numbers

  24. Astrophysics
    • one year ago
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    X cannot be -2 and 2

  25. anonymous
    • one year ago
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    oh.

  26. Astrophysics
    • one year ago
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    \[D: x \in \mathbb{R}: x \neq \pm 2\]

  27. Astrophysics
    • one year ago
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    It's all real except for + and - 2

  28. anonymous
    • one year ago
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    ok:) Thanks :)

  29. Astrophysics
    • one year ago
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    \[R: y \in \mathbb{R}: y \le - \frac{ 3 }{ 4 } ~or~y>0\]

  30. anonymous
    • one year ago
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    So going back to the horizontal asym., I sort of get it know : http://www.purplemath.com/modules/asymtote2.htm

  31. anonymous
    • one year ago
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    Thanks:)

  32. anonymous
    • one year ago
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    so it would be 3/x^2?

  33. Astrophysics
    • one year ago
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    y=0

  34. anonymous
    • one year ago
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    Thanks so much:)))

  35. Astrophysics
    • one year ago
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    and if the degree in the numerator was greater you'd have no horizontal asymptote, so like (x^3+2)/(x^2+3) no HA

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