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chrisdbest

  • one year ago

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  1. chrisdbest
    • one year ago
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    Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -7).

  2. chrisdbest
    • one year ago
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    @pooja195

  3. chrisdbest
    • one year ago
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    @Nnesha @Preetha

  4. jdoe0001
    • one year ago
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    so.. the vertex is a the origin the focus point is a 0. -7 care to draw those two in a cartesian plane?|dw:1437175007773:dw|

  5. chrisdbest
    • one year ago
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    Vertex is at origin, and focus point is at 0, -7 like you said

  6. jdoe0001
    • one year ago
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    ok... so what are the coordinates for the origin?

  7. chrisdbest
    • one year ago
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    Isn't it 0,0?

  8. jdoe0001
    • one year ago
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    yeap so |dw:1437175183982:dw| so.... if the parabola makes a U-turn at the 0,0 and the focus is below the vertex that means the parabola is |dw:1437175290874:dw|

  9. jdoe0001
    • one year ago
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    so is a parabola that opens vertically meaning the "x" term is the one that's squared so \(\bf \begin{array}{llll} (y-{\color{blue}{ k}})^2=4{\color{purple}{ p}}(x-{\color{brown}{ h}}) \\ (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\\ {\color{purple}{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\ \quad \\\\ \quad \\ \textit{so we'd need to use this form }(x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\)

  10. jdoe0001
    • one year ago
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    now... what is the distance "p", that is, from the vertex to the focus point?

  11. chrisdbest
    • one year ago
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    hmmm

  12. jdoe0001
    • one year ago
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    |dw:1437175751363:dw|

  13. chrisdbest
    • one year ago
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    Wouldn't that be 7?

  14. jdoe0001
    • one year ago
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    yeap so p is 7 the parabola is going downards, that means "p" is negative so "p" is really negative 7 or -7 so now you know what "p" is, and where the vertex is at, that is what "h" and "k" are so just plug them in at \(\begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 0}},{\color{blue}{ 0}})\\ {\color{purple}{ 7}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\) and simplify and solve for "y" :)

  15. jdoe0001
    • one year ago
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    well... negative 7 rather \(\begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 0}},{\color{blue}{ 0}})\\ {\color{purple}{ -7}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\)

  16. chrisdbest
    • one year ago
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    Ohk, yeah, I meant down 7

  17. chrisdbest
    • one year ago
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    So what do I do next?

  18. jdoe0001
    • one year ago
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    you shall... plug in those values simplify and solve for "y" :)

  19. chrisdbest
    • one year ago
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    what values?

  20. chrisdbest
    • one year ago
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    So y^2 = -28x?

  21. chrisdbest
    • one year ago
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    @jdoe0001

  22. jdoe0001
    • one year ago
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    \(\large \begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 0}},{\color{blue}{ 0}})\\ {\color{purple}{ -7}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\) notice.... you have all the values you need, just plug them in and solve for "y"

  23. chrisdbest
    • one year ago
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    so what's h?

  24. jdoe0001
    • one year ago
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    look at the colors, h,k are just the vertex coordinates

  25. chrisdbest
    • one year ago
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    ok

  26. chrisdbest
    • one year ago
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    Then what does x equal?

  27. chrisdbest
    • one year ago
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    Ohhh. wait I think I see

  28. chrisdbest
    • one year ago
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    My answer choices are... A. \[y = -\frac{ 1 }{ 7 }x^2\] B. \[y^2 = -7x\] C. \[y=-\frac{ 1 }{ 28 }x^2\] D. \[y^2 = -28x\]

  29. chrisdbest
    • one year ago
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    So I would choose D?

  30. chrisdbest
    • one year ago
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    OMG!!! It's not D!!!

  31. dumbcow
    • one year ago
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    Answer is C note the parabola opens down, so its of the form y = x^2 the "y" is squared only if parabola is sideways or opens left/right

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