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znimon Group Title

How do I find the domain of this function?

  • 2 years ago
  • 2 years ago

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  1. znimon Group Title
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    \[g(x) \sqrt{x^2 - 3x - 40}\]

    • 2 years ago
  2. znimon Group Title
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    I have \[\sqrt{(x-8)(x+5)} \]

    • 2 years ago
  3. znimon Group Title
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    \[(x-8)(x+5) \ge 0\]

    • 2 years ago
  4. znimon Group Title
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    The original problem is at the top.

    • 2 years ago
  5. Yahoo! Group Title
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    Draw a number line and see for it

    • 2 years ago
  6. znimon Group Title
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    I don't know what x can't be though

    • 2 years ago
  7. znimon Group Title
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    I know the answer but I don't know how to get the answer.

    • 2 years ago
  8. Yahoo! Group Title
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    (-infinity , 5] U [8 , infinity)

    • 2 years ago
  9. znimon Group Title
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    yeah but how do I find that?

    • 2 years ago
  10. Yahoo! Group Title
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    just do trial and error method...)

    • 2 years ago
  11. Yahoo! Group Title
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    (x−8)(x+5)≥0

    • 2 years ago
  12. Yahoo! Group Title
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    luk for the vlue of x which satisfies this...

    • 2 years ago
  13. znimon Group Title
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    There must be a logical way to find the answer, right?

    • 2 years ago
  14. ganeshie8 Group Title
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    (x-8)(x+5) >= 0 its a parabola, right ?

    • 2 years ago
  15. znimon Group Title
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    I could input it into my graphing calculator but other wise I don't know how to tell if its a parabola or not.

    • 2 years ago
  16. ganeshie8 Group Title
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    y = ax^2 + bx + c is a parabola

    • 2 years ago
  17. ganeshie8 Group Title
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    any quadratic function gives a parabola graph

    • 2 years ago
  18. znimon Group Title
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    ok

    • 2 years ago
  19. ganeshie8 Group Title
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    can you find the x intercepts ?

    • 2 years ago
  20. ganeshie8 Group Title
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    once you find x-intercepts, you would know where the parabola dips into x axis (goes negative)

    • 2 years ago
  21. ganeshie8 Group Title
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    |dw:1346519548989:dw|

    • 2 years ago
  22. znimon Group Title
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    Oh ya that would be the point at which the g(x) function is 0, so its x - 8 = 0, x = 8, x+ 5 = 0, x=-5, and since its a parabola the graph only passes through zero twice. Nice dude, so the domain is (-infinity, -5] U [8, +infinity]

    • 2 years ago
  23. ganeshie8 Group Title
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    yeah nice you got it :)

    • 2 years ago
  24. znimon Group Title
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    @ganeshie8

    • 2 years ago
  25. znimon Group Title
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    Awesome, thanks again. Is there a way I could recognize any graph's shape or at least some and would you recommend memorizing those?

    • 2 years ago
  26. ganeshie8 Group Title
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    if the degree is 2, then its a parabola

    • 2 years ago
  27. ganeshie8 Group Title
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    finding domains for other polynomials is bit tricky... as they may dip into x-axis multiple times

    • 2 years ago
  28. znimon Group Title
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    okay

    • 2 years ago
  29. ganeshie8 Group Title
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    do you knw sketching polynomilas (end behiavior thingy )

    • 2 years ago
  30. znimon Group Title
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    never heard of it

    • 2 years ago
  31. ganeshie8 Group Title
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    end behavior helps you sketch ANY polynomial freehand

    • 2 years ago
  32. znimon Group Title
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    Thanks dude, I'll look it up

    • 2 years ago
  33. ganeshie8 Group Title
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    process is basically : 1) find x intercepts 2) using end-behavior sketch the graph by hand

    • 2 years ago
  34. ganeshie8 Group Title
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    its easy concept and worth learning... im sure you wil get hang of it in few minutes....... good luck :)

    • 2 years ago
  35. znimon Group Title
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    why is it that a polynomial with a degree of two is a parabola?

    • 2 years ago
  36. znimon Group Title
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    Thanks again.

    • 2 years ago
  37. ganeshie8 Group Title
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    uhh il have to think...... idk exactly... i think it has something to do with local maximum/minimum + increasing/decreasing thingy. we need to turn to calculus for proper understanding. im also learning... . so im no good for answering this :(

    • 2 years ago
  38. ganeshie8 Group Title
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    @mukushla @eliassaab @experimentX

    • 2 years ago
  39. znimon Group Title
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    Oh, I have enough to learn for now so it doesn't matter to me. I bet I wouldn't understand the explanation if it involves terms I've never had before haha.

    • 2 years ago
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