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waheguru

  • 3 years ago

This is a quadratics question and can someone help me graph this

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  1. waheguru
    • 3 years ago
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  2. abb0t
    • 3 years ago
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    If there's \(x^2\) it's a parabola.

  3. waheguru
    • 3 years ago
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    I know but How Do I graph It

  4. waheguru
    • 3 years ago
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    How do I make a proper table of values?

  5. waheguru
    • 3 years ago
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    Do I just guess or is there a way of finding good numbers for x

  6. abb0t
    • 3 years ago
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    You could make a table and pick some points. Say, -2,-1,0,1,2 ?

  7. waheguru
    • 3 years ago
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    Ill try that again but last time i did it i was not getting a curve

  8. waheguru
    • 3 years ago
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    The values I did were -20 -10 0 10 20

  9. waheguru
    • 3 years ago
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    For example For -10 i am getting -9 for y and for 10 i am getting 7.6 for y. This means the graph is not symmetrical

  10. UnkleRhaukus
    • 3 years ago
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    that just means its not symmetrical about the y axis choose x=0,20,40,60,80,100,120 you'll find it symmetrical about x=60

  11. anonymous
    • 3 years ago
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    you are looking at the graph

  12. waheguru
    • 3 years ago
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    I have another question. SO this was not symmetrical along the y axis and when I was making my table of values, I was not able to get a cororect parabola. How can I fnid out wether I should make the x values all positive or negative so I do get the symmetrical point in my table of values?

  13. goformit100
    • 3 years ago
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    |dw:1364746001591:dw|

  14. waheguru
    • 3 years ago
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    Oh, We jsut learned quadratic relations and we have not come accross this formula i guess we will learn this soon

  15. waheguru
    • 3 years ago
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    Thanks for he help

  16. anonymous
    • 3 years ago
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    it says "use graphing technology" so you are allowed to cheat http://www.wolframalpha.com/input/?i=-x^2%2F144%2B5%2F6x++domain+0..120

  17. UnkleRhaukus
    • 3 years ago
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    you can find the x coordinate of the vertex of the parabola by x=-b/2a

  18. anonymous
    • 3 years ago
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    i meant multiply by 144

  19. anonymous
    • 3 years ago
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    \[-\frac{1}{144}x^2+\frac{5}{6}x=0\] multiply by -144 to get \[x^2-120x=0\] factor as \[x(x-120)=0\]

  20. anonymous
    • 3 years ago
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    gives the two zero highest point is smack dab in the middle

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