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anonymous

  • one year ago

The vertex of x2 + 2x - 4y + 9 = 0 A. (-1,2) B. (1,2) C. (-1,-3) D. (1,-3)

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  1. anonymous
    • one year ago
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    Im not really sure how to do this problem

  2. anonymous
    • one year ago
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    -b/2a

  3. UsukiDoll
    • one year ago
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    it's one of those shortcut formulas When working with the vertex form of a quadratic equation \[h = \frac{-b}{2a}\] and k=f(h) . The a and b portions come from \[ax^2+bx+c \]

  4. anonymous
    • one year ago
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    Ok look first let us get the y to a side alone so you get: y=x^2/4 + x/2 + 9/2 right?

  5. anonymous
    • one year ago
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    so plus 4y?

  6. anonymous
    • one year ago
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    The to get the vertex we should find the derivative of y and equate it to 0 correct?

  7. UsukiDoll
    • one year ago
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    derivatives have nothing to do with finding the vertex.

  8. UsukiDoll
    • one year ago
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    derivatives is in all levels of Calculus... we're just dealing with quadratic equations... which is College Algebra or even lower.

  9. UsukiDoll
    • one year ago
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    get this equation in this form \[y=ax^2+bx+c \]

  10. anonymous
    • one year ago
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    4y=x^2+2x+9

  11. anonymous
    • one year ago
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    like that?

  12. UsukiDoll
    • one year ago
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    yeah let me double check this for a minute.

  13. UsukiDoll
    • one year ago
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    \[x^2 + 2x - 4y + 9 = 0 \rightarrow x^2+2x+9=4y \] because we are sort of dealing with this backwards, but I don't think it should be a problem

  14. UsukiDoll
    • one year ago
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    now divide the entire equation by 4

  15. anonymous
    • one year ago
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    Oh really smartipants let me enlighten you the derivative is the slope of the tan at a specific pt so when you find the derivative of a function and you equate to 0 that means your dealing with the line parallel to the x -axis (slope=0) which is the vertex hahahaha everything works on derivates you clever one I dont advise to tell anybody this again because they'll laugh at you :)

  16. anonymous
    • one year ago
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    @UsukiDoll

  17. UsukiDoll
    • one year ago
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    And I just reported you for that comment ^

  18. anonymous
    • one year ago
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    Dylan are you familiar with derivates or not? So ill continue explaining?

  19. UsukiDoll
    • one year ago
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    and if OP isn't familiar with derivatives, that technique shouldn't be used.

  20. anonymous
    • one year ago
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    i am not familiar with it sorry..

  21. UsukiDoll
    • one year ago
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    See I told you that's a bad idea @joyraheb. And you trying to outsmart me makes it even worse. No one should trust anyone with a lower SS.

  22. anonymous
    • one year ago
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    Hahhahahah report thats your way to defend your mathematics 😂 did I even curse well I'll give you an advise, beware of your false knowledge; it is more dangerous than ignorance. :)

  23. UsukiDoll
    • one year ago
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    Being rude on here counts as getting you reported man... and Dylan already admit to not knowing derivatives, so that technique you are suggesting isn't going to work.

  24. anonymous
    • one year ago
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    y=(x^2/4)+.5+2.25

  25. anonymous
    • one year ago
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    is that is divided by 4?

  26. UsukiDoll
    • one year ago
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    keep your equation in fraction form. Decimals are messy.

  27. UsukiDoll
    • one year ago
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    \[x^2+2x+9=4y \rightarrow \frac{x^2}{4}+\frac{1}{2}x+\frac{9}{4}\]

  28. anonymous
    • one year ago
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    okay i see

  29. UsukiDoll
    • one year ago
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    \[\frac{1}{4}x^2+\frac{1}{2}x+\frac{9}{4} = y\]

  30. Owlcoffee
    • one year ago
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    Any parabola can be represented by the parallelism they posses on either axis of reference (xoy). We can take the expression you posted to the structure: \[y=ax^2+bx+c\] This is also the equation that represents a parabola parallel to the y-axis, so I will, without proof give you the structure of the coordinates that represent the vertice: \[V(-\frac{ b }{ 2a },\frac{ 4ac-b^2 }{ 4a })\] So, taking the conic equation you gave: \[x^2+2x-4y+9=0\] We will isolate the "y" term in order to make the coefficients "a", "b" and "c" evident: \[y=\frac{ 1 }{ 4 }x^2+\frac{ 1 }{ 2 }x+\frac{ 9 }{ 4 }\] And then you use the structure I gave you, to find the coordinates of the vertex.

  31. UsukiDoll
    • one year ago
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    so far so good everyone.

  32. UsukiDoll
    • one year ago
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    so now we let \[a= \frac{1}{4}, b =\frac{1}{2}, c =\frac{9}{4} \] and plug it into \[V(-\frac{ b }{ 2a },\frac{ 4ac-b^2 }{ 4a }) \]

  33. anonymous
    • one year ago
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    how can i plug it in if i don't know what a is?

  34. UsukiDoll
    • one year ago
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    remember our quadratic equation . here is the standard form \[ax^2+bx+c \] and you have this equation \[y=\frac{ 1 }{ 4 }x^2+\frac{ 1 }{ 2 }x+\frac{ 9 }{ 4 }\] so comparing these two equations we have our a b and c

  35. anonymous
    • one year ago
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    but you just said a = the equation and i don't know what a is so i can't plug it in the vertex formula

  36. Owlcoffee
    • one year ago
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    The coefficient of "x^2"

  37. UsukiDoll
    • one year ago
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    please read my previous comment. you can get a, b, and c, by comparing your equation to the standard. you will see that a = 1/4, b =1/2, and c=9/4

  38. anonymous
    • one year ago
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    okay

  39. UsukiDoll
    • one year ago
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    \[ax^2+bx+c \] <- standard form \[\frac{ 1 }{ 4 }x^2+\frac{ 1 }{ 2 }x+\frac{ 9 }{ 4 } \] < - your equation.

  40. UsukiDoll
    • one year ago
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    so a = 1/4, b =1/2, and c=9/4

  41. UsukiDoll
    • one year ago
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    our vertex formula \[V(-\frac{ b }{ 2a },\frac{ 4ac-b^2 }{ 4a }) \] you just have to plug a, b, and c into these formulas, then we have our vertex.

  42. UsukiDoll
    • one year ago
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    let's do the first one together if b = 1/2 and a = 1/4 \[\frac{-b}{2a} \] so we place b =1/2 and a =1/4 in that formula \[\LARGE \frac{-\frac{1}{2}}{2(\frac{1}{4})} \] \[\LARGE \frac{-\frac{1}{2}}{(\frac{2}{4})} \]

  43. anonymous
    • one year ago
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    i got (2,1)

  44. anonymous
    • one year ago
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    wait hold on

  45. anonymous
    • one year ago
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    (1,?) i almost got it

  46. UsukiDoll
    • one year ago
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    well let's go a bit slowly on this . \[\LARGE \frac{-\frac{1}{2}}{(\frac{2}{4})} \] the bottom fraction can be reduced further if we divide 2 on the numerator and denominator on 2/4 we have 1/2 \[\LARGE \frac{-\frac{1}{2}}{(\frac{1}{2})} \] then flip the second fraction and we should have \[\frac{-1}{2} \times \frac{2}{1} \]

  47. anonymous
    • one year ago
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    (1,2)???

  48. anonymous
    • one year ago
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    is that right?

  49. anonymous
    • one year ago
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    i just got confused on plugging it in but then i did it what i think was right and thats the answer i got

  50. UsukiDoll
    • one year ago
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    what's \[\frac{-1}{2} \times \frac{2}{1} \] what's -1 x 2 and what's 2 x 1?

  51. anonymous
    • one year ago
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    -2 and 2

  52. UsukiDoll
    • one year ago
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    yes so we have \[\frac{-2}{2} \] now what's -2 divided by 2

  53. anonymous
    • one year ago
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    -1

  54. UsukiDoll
    • one year ago
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    yes that's the first part of our vertex.

  55. anonymous
    • one year ago
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    (-1,?)

  56. UsukiDoll
    • one year ago
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    \[V(-1,\frac{ 4ac-b^2 }{ 4a }) \] a=1/4, b =1/2, c =9/4

  57. anonymous
    • one year ago
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    2

  58. anonymous
    • one year ago
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    (-1,2)

  59. UsukiDoll
    • one year ago
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    we're lucky because the denominator will just be 1 so we just have the numerator to deal with \[\frac{4}{4} \rightarrow 1 \]

  60. UsukiDoll
    • one year ago
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    \[4ac-b^2 \] is all we need to deal with \[\[4\frac{1}{4}\frac{9}{4}-(\frac{1}{2})^2 \] \]

  61. UsukiDoll
    • one year ago
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    so let's deal with the right part of this formula. what's \[(\frac{1}{2})^2 \rightarrow \frac{1}{2} \times \frac{1}{2} \]

  62. anonymous
    • one year ago
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    1/4

  63. UsukiDoll
    • one year ago
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    yeah. \[\[4\frac{1}{4}\frac{9}{4}-\frac{1}{4}\]

  64. UsukiDoll
    • one year ago
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    so now what's \[4 \times \frac{1}{4} \times \frac{9}{4}\] the fraction parts would be easier to do first

  65. anonymous
    • one year ago
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    its 2.25

  66. anonymous
    • one year ago
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    then minus the .25

  67. anonymous
    • one year ago
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    it equals 2

  68. UsukiDoll
    • one year ago
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    errrrrrrrr please keep your answers in fraction form.. decimals are messy!

  69. anonymous
    • one year ago
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    decimals make more sense to me

  70. anonymous
    • one year ago
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    (-1,2)

  71. UsukiDoll
    • one year ago
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    fractions are easier to deal with though...

  72. UsukiDoll
    • one year ago
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    \[4 \times \frac{1}{4} \times \frac{9}{4} -\frac{1}{4} \] \[4 \times \frac{9}{16} -\frac{1}{4} \] \[\frac{36}{16} -\frac{1}{4} \] reducing the left fraction by dividing 4 on the numerator and denominator \[\frac{9}{4} -\frac{1}{4} \] same numerator so we have 8/4 which is 2 :)

  73. UsukiDoll
    • one year ago
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    yay we got our vertex!

  74. anonymous
    • one year ago
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    thank you!

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