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anonymous

  • one year ago

What is the equation of the quadratic graph with a focus of (2, 0) and a directrix of y = −12?

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  1. jdoe0001
    • one year ago
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    |dw:1437691840404:dw| any ideas where those two fellas are in the graph? can you graph them there?

  2. anonymous
    • one year ago
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    |dw:1437691171977:dw|

  3. anonymous
    • one year ago
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    right?

  4. jdoe0001
    • one year ago
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    yeap we also know that the vertex is equidistant from both meaning, the focus point is away from the vertex, the same amount the directrix is away from the vertex let's call that distance "p" so, that means the vertex is "half-way" between the focus and directrix the parabola opens towards the focus, the focus is above the directrix, in this case the parabola opens upwards then |dw:1437692110933:dw|

  5. jdoe0001
    • one year ago
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    the distance from the focus to the vertex is 7, the distance from the directrix to the vertex is 7 the parabola is opening upwards, so the distance "p" is positive the squared variable is the "x" then now that we know where the vertex coordinates are at, and the "p" distance then \(\large \begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\\ {\color{purple}{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\) plug in your values, and solve for "y"

  6. anonymous
    • one year ago
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    wouldn't the focus be on the x-axis, not the y-axis? since it is (2,0) not (0,2)

  7. jdoe0001
    • one year ago
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    hmmmm hold the mayo.. .maybe .... I did mess up there, lemme recheck that

  8. jdoe0001
    • one year ago
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    |dw:1437692548026:dw|

  9. jdoe0001
    • one year ago
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    anyhow, the vertex is half-way, so that makes it a 0,-6 and yes, that makes the "p" distance to 6 only \(\begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 0}},{\color{blue}{ -6}})\\ {\color{purple}{ 6}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\)

  10. jdoe0001
    • one year ago
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    hmmm wait a sec.. dohh. the x coordinate can't be 0 anyhow hehe one sec

  11. jdoe0001
    • one year ago
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    |dw:1437692768286:dw| \(\begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 2}},{\color{blue}{ -6}})\\ {\color{purple}{ 6}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\)

  12. anonymous
    • one year ago
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    ok, we have the vertex, now what do we do to get the equation for it

  13. jdoe0001
    • one year ago
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    just plug in the values in the parabola equation focus form

  14. anonymous
    • one year ago
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    so it ends up as (x-2)^2 = 4*6(y-(-6))

  15. jdoe0001
    • one year ago
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    yeap or \(\bf (x-{\color{brown}{ 2}})^2=4{\color{purple}{ (6)}}(y-{\color{blue}{ (-6)}})\implies (x-2)^2=24(y+6) \\ \quad \\ (x-2)^2=24y+144\implies \cfrac{(x-2)^2-144}{24}=y\)

  16. anonymous
    • one year ago
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    Is there another way to write that out, because none of the above are answer choices

  17. jdoe0001
    • one year ago
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    I guess one could distribute the denominator, and end up with \(\cfrac{(x-2)^2-144}{24}=y\implies \cfrac{(x-2)^2}{24}-\cfrac{\cancel{144}}{\cancel{24}}=y\implies \cfrac{(x-2)^2}{24}-6=y\)

  18. jdoe0001
    • one year ago
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    you could also, expand the squared binomial, and simplify it with the 144, and then distribute the denominator

  19. anonymous
    • one year ago
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    my anwer choices are x^2-4-35; -x^2+4+35; -1/24x^2+1/6x-35/6; and 1/24x^2-1/6-35/6

  20. jdoe0001
    • one year ago
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    ok one sec

  21. jdoe0001
    • one year ago
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    \(\bf \cfrac{(x-2)^2-144}{24}=y\implies \cfrac{[x^2-4x+4]-144}{24}=y \\ \quad \\ \cfrac{x^2-4x+4-144}{24}=y\implies \cfrac{x^2-4x-140}{24}=y \\ \quad \\ \cfrac{x^2}{24}-\cfrac{4x}{24}-\cfrac{140}{24}=y\implies \cfrac{1}{24}x^2-\cfrac{\cancel{4}}{\cancel{24}}x-\cfrac{\cancel{140}}{\cancel{24}}=y\implies ?\)

  22. anonymous
    • one year ago
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    what?

  23. jdoe0001
    • one year ago
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    any... thing.. that confuses there?

  24. anonymous
    • one year ago
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    what happens to the second and third fractions

  25. jdoe0001
    • one year ago
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    yeah... whatever happens to those? so... simplify them, what do you get for the simplification? or 4/24 and 140/24?

  26. anonymous
    • one year ago
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    oh, ok. The would simplify to 2/12 and 35/8

  27. anonymous
    • one year ago
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    sorry 1/6

  28. jdoe0001
    • one year ago
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    well... 2/12 = 1/6 and.... 140/24 simplifiefs to 35/6 thus \(\bf \cfrac{x^2}{24}-\cfrac{4x}{24}-\cfrac{140}{24}=y\implies \cfrac{1}{24}x^2-\cfrac{\cancel{4}}{\cancel{24}}x-\cfrac{\cancel{140}}{\cancel{24}}=y \\ \quad \\ \cfrac{1}{24}x^2-\cfrac{1}{6}x-\cfrac{35}{6}=y\)

  29. anonymous
    • one year ago
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    oh, ok. Thanks for the help

  30. jdoe0001
    • one year ago
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    yw

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