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anonymous

  • one year ago

What is the equation of the quadratic graph with a focus of (5, −1) and a directrix of y = 1? f(x) = −one fourth (x − 5)2 + 1 f(x) = one fourth (x − 5)2 + 1 f(x) = −one fourth (x − 5)2 f(x) = one fourth (x − 5)2

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  1. freckles
    • one year ago
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    i like to draw first... |dw:1438724043667:dw| the vertex will be halfway between the directrix and the focus

  2. freckles
    • one year ago
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    |dw:1438724120865:dw| what are the coordinates for V

  3. anonymous
    • one year ago
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    (0,5)?

  4. freckles
    • one year ago
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    |dw:1438724184117:dw| but isn't (0,5) on the y-axis

  5. anonymous
    • one year ago
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    idk that's why im asking you i have no idea how to do any of this CX

  6. freckles
    • one year ago
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    so you haven't ever plotted points in your class?

  7. anonymous
    • one year ago
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    i have but i never actually paid attention. im one of those kids that sleep in class tbh . and now im making up for it :/

  8. freckles
    • one year ago
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    (x,y) means we are going to travel horizontal x units (note: the sign of x tells us which way we will travel horizontally) means we are going to travel vertically y units (note: the sign of y tells which way we will travel vertically)

  9. freckles
    • one year ago
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    (0,5) tells us we will go 0 units left or right horizontally tells us we will go 5 units up

  10. anonymous
    • one year ago
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    oh so we would be going (5,0) ? Because you're going 5 units to the right and 0 up ?

  11. freckles
    • one year ago
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    right

  12. freckles
    • one year ago
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    the vertex is (5,0)

  13. anonymous
    • one year ago
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    what do i do with that?

  14. anonymous
    • one year ago
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    p.s im also really bad at graphing even if i was good at algebra CX

  15. freckles
    • one year ago
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    |dw:1438724452936:dw|

  16. anonymous
    • one year ago
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    so (5,-1)?

  17. freckles
    • one year ago
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    (5,-1) is the focus you were given this y=1 is the directrix (5,0) is what you found to be the vertex and we just determined p was negative since the parabola is opened down you can find the other part of p by finding the vertical distance between the vertex and the dirextrix which is 1 so p=-1 since we determined it was going to be negative already \[(x-h)^2=4p(y-k) \\ \text{ divide } 4p \text{ on both sides } \frac{1}{4p}(x-h)^2=y-k \\ \text{ add } k \text{ on both sides } \\ y=\frac{1}{4p}(x-h)^2+k\]

  18. freckles
    • one year ago
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    you have already determined the vertex (h,k) was (5,0) and you have determined p is -1 input values and you are done

  19. anonymous
    • one year ago
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    would x =-1?

  20. anonymous
    • one year ago
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    isn'y y -1 @freckles

  21. anonymous
    • one year ago
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    isn't*

  22. freckles
    • one year ago
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    x and y are variables

  23. freckles
    • one year ago
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    you found (h,k) is (5,0) replace h with 5 replace k with 0 you found p=-1 replace p with -1

  24. anonymous
    • one year ago
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    what is x and y?

  25. anonymous
    • one year ago
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    god im so confused :|

  26. anonymous
    • one year ago
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    or nahh XD im dead

  27. freckles
    • one year ago
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    x and y are variables

  28. freckles
    • one year ago
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    can you replace h and k and p with what we have found?

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