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Janu16

  • one year ago

What is the equation of the quadratic graph with a focus of (4, 3) and a directix of y = 13?

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  1. Janu16
    • one year ago
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    @AlexandervonHumboldt2

  2. AlexandervonHumboldt2
    • one year ago
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    as the focus is below the directrix then the graph is concave down. line of symmetry is x=4. |dw:1443884034227:dw| distance between the focus and directrix is 2 multiplied by the focal length. distance is 2b = -10 => focal length is b = -5 units vertex is (4. 3 - (-5)) => (4. 8) (x^2−h)=4a(y−k) (x−4^)2=4*(−5)(y−8) now simplify all this and get your equation

  3. AlexandervonHumboldt2
    • one year ago
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    woops mistake wait

  4. AlexandervonHumboldt2
    • one year ago
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    this is correct distance between the focus and directrix is 2 multiplied by the focal length. distance is 2b = -10 => focal length is b = -5 units vertex is (4. 3 - (-5)) => (4. 8) (x^2−h)=4b(y−k) (x−4)^2=4*(−5)(y−8) now simplify all this and get your equation

  5. Janu16
    • one year ago
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    so do you simplify both of it?

  6. AlexandervonHumboldt2
    • one year ago
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    not both you simplify this equation (x−4)^2=4*(−5)(y−8)

  7. Janu16
    • one year ago
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    = −one twentieth (x − 4)2 + 8

  8. Janu16
    • one year ago
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    |dw:1443884741339:dw|?

  9. Janu16
    • one year ago
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    @AlexandervonHumboldt2

  10. AlexandervonHumboldt2
    • one year ago
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    yes i think so

  11. AlexandervonHumboldt2
    • one year ago
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    though i could have made a mistake

  12. Janu16
    • one year ago
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    ok these are thhe my options hold on

  13. Janu16
    • one year ago
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    |dw:1443885150607:dw|

  14. AlexandervonHumboldt2
    • one year ago
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    the second one as you see

  15. Janu16
    • one year ago
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    other two options are same as these two but without - 20

  16. AlexandervonHumboldt2
    • one year ago
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    ok

  17. AlexandervonHumboldt2
    • one year ago
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    \(\Huge\color{green}{☺☻}\)

  18. Janu16
    • one year ago
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    Alright thx for the help and do you think you can help with another similar problem plz?

  19. AlexandervonHumboldt2
    • one year ago
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    i will try but i'm not sure i will do it

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