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anonymous

  • 5 years ago

I need to graph y=x^2+3. Do I just plug in #'s for x to find y then graph the points??

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  1. anonymous
    • 5 years ago
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    Yes :)

  2. anonymous
    • 5 years ago
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    K... Im just nervous Im doing something wrong. Are these right?? (0,0) (1,4) (2,7) (-1,4) (-2,7)

  3. anonymous
    • 5 years ago
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    Yes, that's correct, well done :)

  4. anonymous
    • 5 years ago
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    My next one I have to graph (LAST ONE!) is y=lx+3l +2 what does that absolute value sign mean??/

  5. anonymous
    • 5 years ago
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    Basically, what ever value is inside the absolute value sign, when you remove the sign, it is always positive. So if you use x = -45, the value inside the absolute bars is -42, but when you remove the bars, it's 42.

  6. anonymous
    • 5 years ago
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    K..So I got the points (0,5) (1,6) (2,7) (-1,6) (-2,7) are they right?

  7. anonymous
    • 5 years ago
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    Not quite, when you use negative x values, you first need to compute the value inside the absolute bars, then remove them. So in the case of x=-1 you get |x+3|=|-1+3|=|2|=2

  8. anonymous
    • 5 years ago
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    So are they (0,5) (1,6) (2,7) (-1,4) (-2,3) are any of them right?

  9. anonymous
    • 5 years ago
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    All are correct. :)

  10. anonymous
    • 5 years ago
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    Thank YOU!!!

  11. anonymous
    • 5 years ago
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    for \[y =x ^{2}+3\] , the y-intercept should be (0, 3) rather than (0,0).

  12. anonymous
    • 5 years ago
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    Obviously I shouldnt help people when I'm busy with other things, too many mistakes today. ^^ Thanks for finding it.

  13. anonymous
    • 5 years ago
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    i know the feeling - also i'm feeding my 3 month old and typing one handed:)

  14. anonymous
    • 5 years ago
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    That's a challenge..! :)

  15. anonymous
    • 5 years ago
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    777777777777777777777777777777777777777777777777777777777777777777777777

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