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anonymous

  • one year ago

Find the vertex of the graph of the function. f(x) = 4x^2 + 24x + 32

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  1. Jhannybean
    • one year ago
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    Basically you're taking your equation \(y=4x^2+24+32\) that has the quadratic form \(y=ax^2+bx+c\) and converting it into vertex form: \(y=a(x-h)^2+k\)

  2. Jhannybean
    • one year ago
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    So to start, we can immediately see that all the coefficients are divisible by 4. So simplify your function.

  3. anonymous
    • one year ago
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    y=4x2+24+32= y=ax2+bx+c ? = y=a(x−h)2+k right?

  4. anonymous
    • one year ago
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    a=4 b=24 c=32

  5. anonymous
    • one year ago
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    y=a(x−h)2+k = = y=4(x−h)2+k

  6. Jhannybean
    • one year ago
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    You got the idea somewhat, but first lets simplify this :) You'll see what I mean by "fitting the equation into the form" :D

  7. anonymous
    • one year ago
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    yeah, kinda. what i don't get is how to get x&y values to then find h&k

  8. anonymous
    • one year ago
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    ohh and thank you so much for helping me @Jhannybean

  9. Jhannybean
    • one year ago
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    Simplify this: \[\large y=\frac{4}{4}x^2+\frac{24}{4}x+\frac{32}{4}\]

  10. Jhannybean
    • one year ago
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    @DanTheMan99

  11. anonymous
    • one year ago
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    thank you

  12. anonymous
    • one year ago
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    sorry mi wifi was acting up @Jhannybean

  13. anonymous
    • one year ago
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    y=x2+6x+8

  14. anonymous
    • one year ago
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    y=x2+6x+8 y=ax2+bx+c ? y=a(x−h)2+k

  15. anonymous
    • one year ago
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    but how will i get the other values?

  16. Jhannybean
    • one year ago
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    Alright, now that we simplified it to \(y=x^2+6x+8\), we need to use the method of completing the square. Have you learned that method yet?

  17. anonymous
    • one year ago
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    not yet

  18. Jhannybean
    • one year ago
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    In the method of completing the square, we will basically create another quadratic function into the quadratic function we already have, \(y=x^2+6x+8\) It's like an inception of quadratics that will help you find the vertex. :P

  19. Jhannybean
    • one year ago
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    So let's work it out slowly. First set your function equal to zero. \(y=x^2+6x+8=0\) Next, move over your constants to one side of the equation, while leaving your variables on the other side side. \(y=x^2+6x=-8\) Following so far?

  20. anonymous
    • one year ago
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    yes

  21. anonymous
    • one year ago
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    do i get x and y intercepts?

  22. Jhannybean
    • one year ago
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    Once we're dne you'll have your x and y intercepts and you'll see what I mean, just follow me on this one :P Trust!

  23. Jhannybean
    • one year ago
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    Now we're going to create a quadratic on the left hand side of the function by completing the square of the left hand side. Essentially we're going to `finish` the quadratic on the left hand side so it resembles the form \(ax^2+bx+c\) but right now, we've only got \(ax^2+bx\) and THEREFORE need to find ourselves a new \(c\) value.

  24. Jhannybean
    • one year ago
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    \[c=\left(\frac{b}{2}\right)^2 \longrightarrow c=\left(\frac{6}{2}\right)^2 = (3)^2 = 9\]

  25. anonymous
    • one year ago
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    so c = 9

  26. Jhannybean
    • one year ago
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    Yep. So since we're `adding` c to the left side, we have to respectively add it to the right side as well

  27. Jhannybean
    • one year ago
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    \[y=x^2+6x\color{red}{+9}=-8\color{red}{+9}\]

  28. anonymous
    • one year ago
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    y=x2+6x+9=−8+9 y=x2+6x+9=+1

  29. anonymous
    • one year ago
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    ohh ok

  30. Jhannybean
    • one year ago
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    If you're good with simplifying quadratics, you'll see that the left hand side now is a PERFECT square! :o \(x^2+6x+9 \longrightarrow (x+3)^2\)

  31. Jhannybean
    • one year ago
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    Rewrite it. \[y=(x+3)^2=1\]

  32. Jhannybean
    • one year ago
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    Well, you would move the 1 over to fit your new equation to fit the vertex form \(y=a(x+h)^2+k\)

  33. Jhannybean
    • one year ago
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    Therefore \(y=a(x+h)^2+k \longrightarrow y=(x+3)^2-1\)

  34. Jhannybean
    • one year ago
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    Do you see how it works?

  35. Jhannybean
    • one year ago
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    Whoop, I wrote it wrong, \(y=a(x-h)^2+k\)** Typo.

  36. anonymous
    • one year ago
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    y=a(x−h)2+k⟶y=(x+3)2−1

  37. anonymous
    • one year ago
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    ⟶y=(x+3)^2−1

  38. Jhannybean
    • one year ago
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    So we know a positive can be split into 2 negatives, therefore \(y=a(x-h)^2+k \longrightarrow y=(x-(-3))^2-1\)

  39. Jhannybean
    • one year ago
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    So tell me, what are your x and y intercepts? :D

  40. anonymous
    • one year ago
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    y=-1

  41. anonymous
    • one year ago
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    xintercept = (x−(−3))2−1 = 1 = (x−(−3))2

  42. Jhannybean
    • one year ago
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    Im not understanding that.

  43. anonymous
    • one year ago
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    1 = (x−(−3))^2

  44. Jhannybean
    • one year ago
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    \[y=a(x\color{red}{-h})^2\color{blue}{+k} \longrightarrow y=(x-\color{red}{(-3)})^2\color{blue}{-1}\]

  45. Jhannybean
    • one year ago
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    To help you identify them a little better. the part in red is your x-intercept, part in blue is your y intercept

  46. anonymous
    • one year ago
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    thats much better, sorry i was confused

  47. anonymous
    • one year ago
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    \[y=a(x−h)^{2}+k⟶y=(x−(−3))^{2}−1\]

  48. anonymous
    • one year ago
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    wait so what do i do now? the intercepts?

  49. Jhannybean
    • one year ago
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    You're toldto find the vertex of the graph, which I helped you find in the post above :)

  50. anonymous
    • one year ago
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    thank you @Jhannybean , sorry im a little heard-headed but i just reread it and i completely understand

  51. Jhannybean
    • one year ago
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    Awesome! Glad I could help :)

  52. anonymous
    • one year ago
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  53. anonymous
    • one year ago
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    @Jhannybean sorry to bother you again

  54. anonymous
    • one year ago
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    does it mean that the answer would be "b"?

  55. anonymous
    • one year ago
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    im so confused :(

  56. anonymous
    • one year ago
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    use is formula....\[-\frac{ b }{ 2a }\]

  57. anonymous
    • one year ago
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    do i plug in a=1 and b=6??

  58. anonymous
    • one year ago
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    \[y=ax^2+bx+c\]

  59. anonymous
    • one year ago
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    y=(x−(−3))^2−1 -\[\frac{ -3 }{ 1 }\]

  60. anonymous
    • one year ago
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    -(-3/2)

  61. anonymous
    • one year ago
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    right?

  62. anonymous
    • one year ago
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    |dw:1438666779185:dw|

  63. anonymous
    • one year ago
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    3

  64. anonymous
    • one year ago
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    a and b is coefficient

  65. anonymous
    • one year ago
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    all the numberse

  66. anonymous
    • one year ago
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    24/8

  67. anonymous
    • one year ago
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    add a negative sign

  68. anonymous
    • one year ago
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    -3

  69. anonymous
    • one year ago
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    yea

  70. anonymous
    • one year ago
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  71. anonymous
    • one year ago
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    @saseal

  72. anonymous
    • one year ago
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    would it be the -3,-4 or -4,-3

  73. anonymous
    • one year ago
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    lowest point is obviously the y-axis

  74. anonymous
    • one year ago
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    im so lost

  75. dan815
    • one year ago
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    dantheman

  76. dan815
    • one year ago
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    lisnten to me, do u know how transformations work with functions

  77. anonymous
    • one year ago
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    GET A LIFE DAN SRS

  78. anonymous
    • one year ago
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    no

  79. anonymous
    • one year ago
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    sorry

  80. dan815
    • one year ago
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    do you know how transformations work??

  81. anonymous
    • one year ago
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    kind of

  82. dan815
    • one year ago
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    okay how about completing the square

  83. anonymous
    • one year ago
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    f(x-3) = f(x) shifted 3 to the right as example

  84. anonymous
    • one year ago
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    ok i get that

  85. anonymous
    • one year ago
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    if i am subtracing 3 away from what i am inputting at the time, then it must appear later by 3 units

  86. dan815
    • one year ago
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    f(x) = 4x^2 + 24x + 32 the point im trying to get across is we want to write this equation in the form f(x) = a(x-b)^2+c this means the vertex is at (b,c)

  87. anonymous
    • one year ago
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    i know its - 3

  88. dan815
    • one year ago
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    okay forget that BS for now

  89. dan815
    • one year ago
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    do you get why this equation has vertex at b,c though

  90. anonymous
    • one year ago
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    what i dont know is if it goes before or after the -4

  91. dan815
    • one year ago
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    f(x) = a(x-b)^2+c

  92. dan815
    • one year ago
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    no forget all that stuff, we are going to do this logically

  93. dan815
    • one year ago
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    f(x) = a(x-b)^2+c do you get why the vertex for this equation is b,c

  94. dan815
    • one year ago
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    think about the base graph

  95. dan815
    • one year ago
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    y=x^2

  96. dan815
    • one year ago
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    how do you apply transformations to it?

  97. dan815
    • one year ago
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    y=(x-b)^2 would shift it b units to the right

  98. dan815
    • one year ago
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    y=(x-b)^2 + c would shift it b units to the right and c units up

  99. dan815
    • one year ago
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    you can think about it like this too (y-c)=(x-b)^2 y shifted c up and x shifted b right

  100. anonymous
    • one year ago
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    THIS MAN HAS CLEARLY LOST IT

  101. anonymous
    • one year ago
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    sorry my wifi wasn't working

  102. anonymous
    • one year ago
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    okay im following

  103. dan815
    • one year ago
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    do u get why the vertex for y=a(x-b)^2+c is at (b,c) then?

  104. anonymous
    • one year ago
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    yes

  105. dan815
    • one year ago
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    okay then y=4x^2 + 24x + 32 rewrite this in that form

  106. anonymous
    • one year ago
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    y=a(x-b)^2+c y=4(x-24)^2+32

  107. dan815
    • one year ago
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    thats not how it works

  108. dan815
    • one year ago
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    expand the brackets see if u get the right equation

  109. anonymous
    • one year ago
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    y=x^2+6x+8

  110. anonymous
    • one year ago
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    ohhhh

  111. dan815
    • one year ago
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    y=4(x-24)^2+32 =4*(x-24)(x-24)+32 = 4*(x^2-48x+24^2)+32

  112. dan815
    • one year ago
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    its much bigger than before now

  113. dan815
    • one year ago
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    so we have to transform to the form we want properly

  114. anonymous
    • one year ago
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    okay so you did the square

  115. dan815
    • one year ago
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    okay ill show u how to do it this time, u can try to do it later

  116. anonymous
    • one year ago
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    (a+b)^2=a^2+2ab+b^2

  117. anonymous
    • one year ago
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    = 4*(x^2-48x+24^2)+32

  118. dan815
    • one year ago
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    y=4x^2 + 24x + 32 y=4(x^2+6x)+32 ill complete the square in the brackets =4(x^2+6x+9-9)+32 =4((x^2+6x+9)-9)+32 =4*((x+3)^2-9)+32 =4*(x+3)^2+32-9*4 =4*(x+3)^2+32-36 =4*(x+3)^2-4 so the vertex is at (-3,-4)

  119. anonymous
    • one year ago
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    thank you so much

  120. anonymous
    • one year ago
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    uhhhhh finally i understand

  121. dan815
    • one year ago
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    y=4x^2 + 32x + 32 tell me the vertex for this equation and i will believe you

  122. dan815
    • one year ago
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    follow the steps i went through up there, its a very similar question

  123. anonymous
    • one year ago
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    @dan815 whoa you went thru all at

  124. dan815
    • one year ago
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    it looks like a lot of steps because im showing every little change, but normally once u learn the method u do it all in 1 or 2 steps

  125. dan815
    • one year ago
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    u can do these in ur head pretty much

  126. anonymous
    • one year ago
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    yea ikr thats why i said that lol

  127. anonymous
    • one year ago
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    y=4x^2 + 32x + 32 y=4(x^2+8x)+32 would it be that

  128. anonymous
    • one year ago
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    so i can continue

  129. dan815
    • one year ago
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    thats a start now u have to complete the square in those brackets go on

  130. anonymous
    • one year ago
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    y=4x^2 + 32x + 32 y=4(x^2+8x)+32 =4(x^2+8x+9-9)+32 =4((x^2+8x+9)-9)+32 =4*((x+4)^2-9)+32 =4*(x+4)^2+32-9*4 =4*(x+4)^2+32-36 =4*(x+4)^2-4

  131. anonymous
    • one year ago
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    :w

  132. dan815
    • one year ago
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    no

  133. dan815
    • one year ago
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    the question u are asking urself here is how can i get a perfect square out of x^2+8x

  134. dan815
    • one year ago
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    when u see x^2+8x you have to realize that the square has to be (x+4)^2 since this is the only square that will give u a x^2 and 8x now the constant is not there so we add and subtract that the constant part will be 4^2=16 so x^2+8x+16-16

  135. anonymous
    • one year ago
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    so you have to multiply 8 by 2

  136. dan815
    • one year ago
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    u are not multiplying 8 by 2

  137. dan815
    • one year ago
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    u are dividing (8/2) then squaring (8/2)^2 = 16 just so happesn that its also the same as 8*2

  138. anonymous
    • one year ago
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    y=4x^2 + 32x + 32 y=4(x^2+8x)+32 =4(x^2+8x+16-16)+32 =4((x^2+8x+16)-16)+32 =4*((x+4)^2-16)+32 =4*(x+4)^2+32-16*4

  139. dan815
    • one year ago
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    when u have x^2+ax and u want to write it in square form you take half of a as (x+a/2)^2=x^2+ax+(a/2)^2 ^--- term on x works out

  140. dan815
    • one year ago
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    yes now u got it so what is the vertex

  141. anonymous
    • one year ago
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    -4,-4??

  142. dan815
    • one year ago
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    no

  143. dan815
    • one year ago
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    what is the constant term

  144. anonymous
    • one year ago
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    16

  145. dan815
    • one year ago
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    =4*(x+4)^2+32-16*4 simplfy this to a(x-b)^2+c

  146. dan815
    • one year ago
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    =4*(x+4)^2+32-16*4 =4*(x+4)^2+32-64 =??

  147. anonymous
    • one year ago
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    -32

  148. anonymous
    • one year ago
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    =4*(x+4)^2-32

  149. dan815
    • one year ago
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    oaky so what is the vertex?

  150. anonymous
    • one year ago
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    i keep thinking its -4,-3

  151. anonymous
    • one year ago
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    but i don't think it is that

  152. dan815
    • one year ago
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    what is the vertex for this equation =a(x-b)^2+c

  153. anonymous
    • one year ago
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    b,c

  154. dan815
    • one year ago
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    okay so what is the vertex for this =4*(x+4)^2-32

  155. anonymous
    • one year ago
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    -4,-32

  156. dan815
    • one year ago
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    yes

  157. anonymous
    • one year ago
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    wow

  158. dan815
    • one year ago
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    :)

  159. anonymous
    • one year ago
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    that was really hard but now i got it lol

  160. anonymous
    • one year ago
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    only the inside changes sign, its the opposite

  161. anonymous
    • one year ago
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    thank you dan815

  162. anonymous
    • one year ago
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    you're a lifesaver!

  163. dan815
    • one year ago
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    no problem, try some completing the square excercises just to make sure though

  164. dan815
    • one year ago
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    or just do some expanding excercised, that should build your intuition up

  165. anonymous
    • one year ago
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    i tried doing it with analytic geometry method lol

  166. anonymous
    • one year ago
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    thank you

  167. anonymous
    • one year ago
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    works too

  168. dan815
    • one year ago
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    dantheman

  169. dan815
    • one year ago
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    for the record here is a short cut

  170. anonymous
    • one year ago
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    would it be the same for Writing the quadratic function in vertex form. y = x2 + 8x + 18

  171. dan815
    • one year ago
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    f(x) = 4x^2 + 24x + 32 you take the middle number and divide by -2* the first number 24/(-2*4) = -24/8=-3 now plug -3 into equation f(-3)=4*9+24*-3+32=-4 vertex is at (3,f(-3)) = (-3,-4)

  172. anonymous
    • one year ago
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    ohh wow that as much faster

  173. dan815
    • one year ago
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    NO do not use it

  174. dan815
    • one year ago
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    since u do not understand it

  175. dan815
    • one year ago
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    not until you understand completing the square method, then u can derive this

  176. dan815
    • one year ago
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    and then u can derive it one more time after u learn about derivatives

  177. anonymous
    • one year ago
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    yea i told him that stuff earlier lol -b/2a

  178. dan815
    • one year ago
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    f(x)=ax^2+bx+c prove that the vertex is at (-b/2a, ) x_coord=-b/2a y_coord=f(-b/2a) This can be your home work or something

  179. anonymous
    • one year ago
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    thank you

  180. dan815
    • one year ago
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    hint complete square for f(x)=ax^2+bx+c

  181. anonymous
    • one year ago
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    :)

  182. anonymous
    • one year ago
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    how can i rate a qualified helper

  183. anonymous
    • one year ago
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    (5 starts)

  184. anonymous
    • one year ago
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    lol

  185. Jhannybean
    • one year ago
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    Similar to my method without simplifying the equation first

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