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shelby1290

  • one year ago

For each quadratic relation, determine the y-intercept and the axis of symmetry a) y= x^2 + 2 b) y= -x^2 -4

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  1. shelby1290
    • one year ago
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    For part A) I got y=2 because I set x to 0 to get my y-intercept y=x^2+2 y=0^2+2 y=2 (0,2)

  2. shelby1290
    • one year ago
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    I then set my y-int to 0 to get my x y=x^2+2 0=x^2+2 0=1x+2 x=2 (2,0)

  3. shelby1290
    • one year ago
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    Am I doing it right so far? How do i find the axis of symmetry?

  4. anonymous
    • one year ago
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    |dw:1441945461825:dw|

  5. anonymous
    • one year ago
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    |dw:1441945660612:dw|

  6. shelby1290
    • one year ago
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    Do I need to find all of the variables?

  7. anonymous
    • one year ago
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    you can do it analytically, but with the vertex form it is not so easy to see it right away unless you are pretty well-versed with quadratic equations with given vertex-form you can solve for the axis of symmetry by first converting the equation into the standard form \(f(x) = ax^2+bx+c\) and the formula for the axis of symmetry is \(\large x=\frac{-b}{2a} \) now you will notice that symmetry pertains to the relationship of one half to the other half of the graph and their similarity

  8. anonymous
    • one year ago
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    if you looked at your other notes, you will probably also notice that the axis of symmetry is your x-value in your vertex

  9. anonymous
    • one year ago
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    |dw:1441946631528:dw|

  10. anonymous
    • one year ago
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    the neat thing about it in vertex form is that it is already laid out to you where your axis of symmetry is.

  11. anonymous
    • one year ago
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    but if you want to just punish yourself by doing more work, let us see if it is truly the formula I gave you for the axis of symmetry based on the standard form of quadratic equation |dw:1441946825142:dw|

  12. anonymous
    • one year ago
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    let us provide a clear understanding of what x-intercept and y-intercepts are |dw:1441947158858:dw|

  13. shelby1290
    • one year ago
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    Okay I'm following......

  14. anonymous
    • one year ago
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    |dw:1441947258243:dw|

  15. shelby1290
    • one year ago
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    mhm

  16. anonymous
    • one year ago
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    |dw:1441947409829:dw|

  17. anonymous
    • one year ago
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    |dw:1441947459955:dw|

  18. anonymous
    • one year ago
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    |dw:1441947486217:dw|

  19. anonymous
    • one year ago
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    |dw:1441947550094:dw|

  20. anonymous
    • one year ago
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    now that we have that clear note that your axis of symmetry is not the the same as your intercepts they may have the same value on some circumstances, but not always not the case

  21. anonymous
    • one year ago
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    |dw:1441947737731:dw|

  22. shelby1290
    • one year ago
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    So, can the axis of symmetry basically be thought of as a midpoint?

  23. anonymous
    • one year ago
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    somewhat but not quite because midpoint is a very precise definition

  24. anonymous
    • one year ago
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    |dw:1441947959255:dw|

  25. anonymous
    • one year ago
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    |dw:1441947998746:dw|

  26. anonymous
    • one year ago
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    difficult and not possible, correct?

  27. shelby1290
    • one year ago
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    |dw:1441948105567:dw|

  28. anonymous
    • one year ago
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    |dw:1441948201968:dw|

  29. shelby1290
    • one year ago
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    But it can kinda be anywhere around the perimeter of the circle...we'll just need coordinates

  30. shelby1290
    • one year ago
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    |dw:1441948232996:dw|