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anonymous

  • one year ago

Finding the domain and range

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  1. anonymous
    • one year ago
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    I need help finding the domian and range for my graph for y=3(x-2)^2+1

  2. Z4K4R1Y4
    • one year ago
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    this should help: http://www.purplemath.com/modules/fcns2.htm

  3. anonymous
    • one year ago
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    @Nnesha hey

  4. anonymous
    • one year ago
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    @Nnesha could you help?

  5. anonymous
    • one year ago
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    Hey Nnesha you helped me last time. I was wondering if you were familiar with this?

  6. Nnesha
    • one year ago
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    hey blue alright so are u allowed to graph it ?

  7. anonymous
    • one year ago
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    yeah i have to graph it ( i did already) now i have to find the domain and range

  8. Nnesha
    • one year ago
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    domain of parabola is always going to be negative inf to positive inf|dw:1433021151849:dw|

  9. Nnesha
    • one year ago
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    |dw:1433021252047:dw|we don't know how longer its going to be so that's why domain is \[(- ∞, ∞)\]

  10. Nnesha
    • one year ago
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    now range is y value so |dw:1433021354845:dw| and range start from the vertex(y point) to positive ∞

  11. Nnesha
    • one year ago
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    the red line shows starting point of range

  12. anonymous
    • one year ago
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    okay so the range is (positive inf,1)

  13. anonymous
    • one year ago
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    no its

  14. Nnesha
    • one year ago
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    starting point come first (start, end) so you don't know where its going to end so positive infinity

  15. anonymous
    • one year ago
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    [1,positive inf)

  16. Nnesha
    • one year ago
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    yep right

  17. anonymous
    • one year ago
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    okay great I understand it now. one more question

  18. Nnesha
    • one year ago
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    okay

  19. Nnesha
    • one year ago
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    |dw:1433021812495:dw|

  20. anonymous
    • one year ago
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    okay yeah so i list those points?

  21. Nnesha
    • one year ago
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    parabola graph isn't crossing x-axis so no x-intercept is there any point where line cross y -axis ?

  22. anonymous
    • one year ago
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    No

  23. Nnesha
    • one year ago
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    mhm yes there is a point where line cross y-axis if you r using desmos caluclator zoomout a little bit so you can see

  24. anonymous
    • one year ago
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    okay one sec

  25. anonymous
    • one year ago
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    my vertex for this problem is (2,1) my graph looks different from yours

  26. Nnesha
    • one year ago
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    oh yeah it's that's just an example (weird drawing ik ik ;D)

  27. Nnesha
    • one year ago
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    |dw:1433022493243:dw| find this blue point where line cross y-axis

  28. anonymous
    • one year ago
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    LOL okay

  29. anonymous
    • one year ago
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    |dw:1433022652761:dw| my graph literally shows like this because the points that i plotted were (1,4) and (3,4)

  30. Nnesha
    • one year ago
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    plotted point why did you plot points ?

  31. Nnesha
    • one year ago
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    http://prntscr.com/7b9y0p this is your graph do you see the point where line cross y-axis between 10 and 15 you need that point for y-intercept zoomout a little bit to see :-)

  32. anonymous
    • one year ago
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    okay i see. im doing this by hand, so my graph only goes up to 5 lol. oops

  33. anonymous
    • one year ago
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    Okay now lastly

  34. Nnesha
    • one year ago
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    ohhh okay so then you have to plug in 0 for x bec when line cross y-axis x value is 0

  35. Nnesha
    • one year ago
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    ^^^ to find y-intercept b hand

  36. anonymous
    • one year ago
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    the main question is for me to compare f(x)= a(x-h)^2 +k to f(x) x^2. i explained that they are basically the same its just that the h and k show us how left/right/up/down the curve has shifted. does that make sense?

  37. Nnesha
    • one year ago
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    y = x^2 is a basic form of quadratic function it shows the shape of parabola yes you are right the difference f(x)= a(x-h)^2 +k is a vertex form equation where h represent horizontal shift and k is for vertical shift where a is tell how how wider or Skinner the graph is going to be

  38. Nnesha
    • one year ago
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    s*

  39. Nnesha
    • one year ago
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    in other words f(x)=x^2 is a parent function

  40. anonymous
    • one year ago
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    Okay great YAY! I feel smart Thank you so much for helping me better understand, I wish I could repay you. Your not a qualified helper right? :-(

  41. Nnesha
    • one year ago
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    lol glad to hear that :-) yes i'm not QH and btw you didn't ask QH question :-)

  42. anonymous
    • one year ago
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    Well thank you so much for taking the time out to help me, I really appreciate.

  43. Nnesha
    • one year ago
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    my pleasure :-) thanks!

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