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znimon

  • 3 years ago

How do I find the domain of this function?

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  1. znimon
    • 3 years ago
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    \[g(x) \sqrt{x^2 - 3x - 40}\]

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

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

  4. znimon
    • 3 years ago
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    The original problem is at the top.

  5. Yahoo!
    • 3 years ago
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    Draw a number line and see for it

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

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

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

  9. znimon
    • 3 years ago
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    yeah but how do I find that?

  10. Yahoo!
    • 3 years ago
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    just do trial and error method...)

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

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

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

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

  15. znimon
    • 3 years ago
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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.

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

  17. ganeshie8
    • 3 years ago
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    any quadratic function gives a parabola graph

  18. znimon
    • 3 years ago
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    ok

  19. ganeshie8
    • 3 years ago
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    can you find the x intercepts ?

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

  21. ganeshie8
    • 3 years ago
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    |dw:1346519548989:dw|

  22. znimon
    • 3 years ago
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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]

  23. ganeshie8
    • 3 years ago
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    yeah nice you got it :)

  24. znimon
    • 3 years ago
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    @ganeshie8

  25. znimon
    • 3 years ago
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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?

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

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

  28. znimon
    • 3 years ago
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    okay

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

  30. znimon
    • 3 years ago
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    never heard of it

  31. ganeshie8
    • 3 years ago
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    end behavior helps you sketch ANY polynomial freehand

  32. znimon
    • 3 years ago
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    Thanks dude, I'll look it up

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

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

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

  36. znimon
    • 3 years ago
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    Thanks again.

  37. ganeshie8
    • 3 years ago
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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 :(

  38. ganeshie8
    • 3 years ago
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    @mukushla @eliassaab @experimentX

  39. znimon
    • 3 years ago
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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.

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