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whatevs

  • 3 years ago

The graph of y = x2 is shown below. Using complete sentences, explain how the graph of y = –2(x + 3)2 + 1 would differ from this parabola?

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  1. whatevs
    • 3 years ago
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    sorry..not a good artist..the vertex is (0,0)

  2. whatevs
    • 3 years ago
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    |dw:1322072661463:dw|

  3. karatechopper
    • 3 years ago
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    r u doing absolute valuee functions/equations?

  4. whatevs
    • 3 years ago
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    yes..I need hlep with this

  5. karatechopper
    • 3 years ago
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    dont worry i just learned this a couple of days ago i was frustrated too but i can help in this now:)

  6. whatevs
    • 3 years ago
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    thank you

  7. karatechopper
    • 3 years ago
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    r the parentheses supposed to be the absolute value signs? and is x2 x times 2?

  8. luckey
    • 3 years ago
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    the graph of y=-2(x+3)^2+1 will have curvature downwards and its vertex is at \[(1/\sqrt{2}-3.0)\]

  9. karatechopper
    • 3 years ago
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    woah...^

  10. whatevs
    • 3 years ago
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    yes, and x2 is x^2

  11. whatevs
    • 3 years ago
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    thank you guyz

  12. karatechopper
    • 3 years ago
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    ok

  13. karatechopper
    • 3 years ago
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    well i luckey has given answer so good job

  14. whatevs
    • 3 years ago
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    U still tried

  15. karatechopper
    • 3 years ago
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    yea im proud of my self:) ROBIN AWAY!

  16. karatechopper
    • 3 years ago
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    thnx

  17. myininaya
    • 3 years ago
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    \[y=-2(x+3)^2+1 \text{ has vertex } (-3,1)\]

  18. myininaya
    • 3 years ago
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    \[y=a(x-h)^2+k \text{ has vertex } (h,k)\] if a>0 then the parabola is concave up if a<0 then the parabola is concave down

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