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anonymous

  • one year ago

The functions f(x) = –(x + 4)2 + 2 and g(x) = (x − 2)2 − 2 have been rewritten using the completing-the-square method. Is the vertex for each function a minimum or a maximum? Explain your reasoning for each function.

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  1. SolomonZelman
    • one year ago
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    If your quadratic has a negative leading coefficient, then it opens down, and its vertex is the aboslute maximum. If your quadratic has a positive leading coefficient, then it opens up, and its vertex is the absolute minimum.

  2. SolomonZelman
    • one year ago
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    that is the rule.....

  3. anonymous
    • one year ago
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    Ok so the first one would be that it opens down correct?

  4. anonymous
    • one year ago
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    thank-you:) Can I ask anither?

  5. SolomonZelman
    • one year ago
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    yes, the first one opens down. Right;)

  6. SolomonZelman
    • one year ago
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    And the second one opens where?

  7. SolomonZelman
    • one year ago
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    (and yes, you can ask another question right here, after you feel that you are done with this question.)

  8. anonymous
    • one year ago
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    It opens us since it's poitive

  9. SolomonZelman
    • one year ago
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    yes

  10. SolomonZelman
    • one year ago
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    ok i guess you are done with this question... r u?

  11. SolomonZelman
    • one year ago
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    do you have any questions about this question ?

  12. anonymous
    • one year ago
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    so the first one, is opening down w/ the vertex at an absolute maximum? I'm just checking

  13. SolomonZelman
    • one year ago
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    yes

  14. SolomonZelman
    • one year ago
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    the vertex in function 1 is absolute maximum

  15. anonymous
    • one year ago
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    Ok thank-you:)

  16. anonymous
    • one year ago
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    Suppose C and D represent two different school populations where C > D and C and D must be greater than 0. Which of the following expressions is the largest? Explain why. Show all work necessary. (C + D)2 2(C + D) C2 + D2 C2 − D2

  17. SolomonZelman
    • one year ago
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    yw

  18. anonymous
    • one year ago
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    i don't understand this one

  19. SolomonZelman
    • one year ago
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    you don't understand the question, or don't know how to solve it?

  20. anonymous
    • one year ago
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    I don't understand how to solve it

  21. SolomonZelman
    • one year ago
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    oh....ok...

  22. SolomonZelman
    • one year ago
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    I will also add, that C and D must be integers (since you can't have 3.5 people of anything like that).... do you agree with me?

  23. anonymous
    • one year ago
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    yes. Yes I do

  24. SolomonZelman
    • one year ago
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    Now, C is at least 1. D is at least 2. (So, C+D is at least 3) Which is the largest? Explain why. Show all work necessary. (C + D)² 2(C + D) C² + D² C² − D²

  25. SolomonZelman
    • one year ago
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    (C+D)² when C=1 & D=@ is going to be (1+2)²=3²=9 2(C+D) 2(1+2)=2(3)=6

  26. SolomonZelman
    • one year ago
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    2(C+D) is a linear growth, and (C+D)² is a "quadratic" growth. And the bigger "C+D" we have the greater will (C+D) get when compared to (C+D) that will not grow even nearly as rapidly.

  27. SolomonZelman
    • one year ago
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    the last option is going to be obviously smaller than or greater than the 3rd option ?

  28. SolomonZelman
    • one year ago
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    can you tell me?

  29. anonymous
    • one year ago
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    so the answer would be c2+d2

  30. SolomonZelman
    • one year ago
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    hold on...

  31. SolomonZelman
    • one year ago
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    pliz answer my last question...

  32. SolomonZelman
    • one year ago
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    (you are trying to jump ahead to quickly:o)

  33. anonymous
    • one year ago
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    smalleer :) Sorry

  34. SolomonZelman
    • one year ago
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    yes (no need for apologies).

  35. SolomonZelman
    • one year ago
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    So, the last option falls off... and in competition is the 1st option (C+D)² which is larger than the 2nd option (as i showed), and the 3rd option (which is obviously larger than the 4th option).

  36. SolomonZelman
    • one year ago
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    ((remember D must be greater than C, and both C and D are natural numbers)) C²+D² vs (C+D)² DIFFERENCE ----------------------------------------------------------- C=1 & D=2 1²+2²=5 vs (1+2)²=9 (C+D)² exceeds by 4 ----------------------------------------------------------- C=1 & D=3 1²+3²=10 vs (1+3)²=16 (C+D)² exceeds by 6 ----------------------------------------------------------- C=2 & D=3 2²+3²=13 vs (2+3)²=25 (C+D)² exceeds by 12 ----------------------------------------------------------- C=2 & D=4 2²+4²=18 vs (2+4)²=36 (C+D)² exceeds by 18 (& twice)

  37. SolomonZelman
    • one year ago
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    The greater values for C and D we pick, the more (C+D)² exceeds, (outmatches if you will) the C²+D². So which one is the largest?

  38. anonymous
    • one year ago
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    )c+d)^2

  39. SolomonZelman
    • one year ago
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    Yes, (C+D)² is the largest, and this it is the answer you need:)

  40. SolomonZelman
    • one year ago
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    Any questions about this problem?

  41. anonymous
    • one year ago
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    No. :)

  42. SolomonZelman
    • one year ago
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    OK.....

  43. anonymous
    • one year ago
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    Three functions are given below: f(x), g(x), and h(x). Explain how to find the axis of symmetry for each function, and rank the functions based on their axis of symmetry (from smallest to largest). f(x) g(x) h(x) f(x) = 3(x + 4)2 + 1 g(x) = 2x2 − 16x + 15

  44. SolomonZelman
    • one year ago
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    ok, this is a last question for this post. Alright>

  45. anonymous
    • one year ago
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    Yea:)

  46. SolomonZelman
    • one year ago
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    h(x) = ?

  47. anonymous
    • one year ago
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    h(x): graph parabola (1,-3) (3,5) (-1,5)

  48. SolomonZelman
    • one year ago
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    for h(x), if you were to graph the points:|dw:1435801328063:dw|

  49. SolomonZelman
    • one year ago
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    you can tell that (1,-3) is the vertex od the parabola.

  50. anonymous
    • one year ago
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    ys it is the vertex

  51. SolomonZelman
    • one year ago
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    And that way we know that \(\Large y=\color{red}{\rm a}(x-1)^2-3\) part.

  52. SolomonZelman
    • one year ago
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    we know everything besides the scale factor (or the coefficient) we will use.

  53. SolomonZelman
    • one year ago
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    (btw, if you don't undertsnad something completely, you are welcome to ask anything that you would like)

  54. anonymous
    • one year ago
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    Ok. I understand everything so far. No worries:)

  55. SolomonZelman
    • one year ago
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    Ok, yes... so now we need to use one of your point. We know that this parabola should satisfy the point (3,5)

  56. SolomonZelman
    • one year ago
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    don't forget that this parabola that we are finding is the h(x)

  57. SolomonZelman
    • one year ago
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    So, \(\large\color{black}{ \displaystyle h(x)=\color{red}{\rm a}(x-1)^2-3 }\) we know that point (3,5) should be on the parabola. So: \(\large\color{black}{ \displaystyle \color{blue}{5}=\color{red}{\rm a}(\color{blue}{3}-1)^2-3 }\)

  58. SolomonZelman
    • one year ago
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    Can you solve for a, please?

  59. anonymous
    • one year ago
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    ok hold on

  60. anonymous
    • one year ago
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    a=1/5

  61. anonymous
    • one year ago
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    5=a(4)-3 5=a1 a=1/5

  62. anonymous
    • one year ago
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    hello?

  63. anonymous
    • one year ago
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    @SolomonZelman is it g(x), f(x), and then h(x) ? I'm sorry

  64. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \color{blue}{5}=\color{red}{\rm a}(\color{blue}{3}-1)^2-3 }\) \(\large\color{black}{ \displaystyle \color{blue}{5}=\color{red}{\rm a}(2)^2-3 }\) \(\large\color{black}{ \displaystyle \color{blue}{5}=4\color{red}{\rm a}-3 }\) \(\large\color{black}{ \displaystyle 2=4\color{red}{\rm a} }\) a=½

  65. SolomonZelman
    • one year ago
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    this a that we found, we found for h(x).....

  66. SolomonZelman
    • one year ago
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    So, \(\large\color{black}{ \displaystyle f(x) = 3(x + 4)^2 + 1 }\) \(\large\color{black}{ \displaystyle g(x) = 2x^2 − 16x + 15 }\) \(\large\color{black}{ \displaystyle h(x) = \frac{1}{2}(x -1)^2 -3 }\) (we foudn a=1/2)

  67. anonymous
    • one year ago
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    I see, I see

  68. SolomonZelman
    • one year ago
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    and x-axis of summetry (in this case) is just the x-coordinate of the vertex.

  69. SolomonZelman
    • one year ago
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    just "axis of summetry" (not "x-axis of summetry")

  70. SolomonZelman
    • one year ago
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    you need to complete the square for g(x)

  71. anonymous
    • one year ago
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    So, h(x) axis of symmetry would be 1?

  72. anonymous
    • one year ago
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    @SolomonZelman I have no idea how to do that

  73. SolomonZelman
    • one year ago
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    yes for h(x) it is 1

  74. SolomonZelman
    • one year ago
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    can you identify the axis of summetry for f(x) ?

  75. anonymous
    • one year ago
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    -4

  76. SolomonZelman
    • one year ago
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    yes

  77. SolomonZelman
    • one year ago
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    Now, for g(x). \(\large\color{black}{ \displaystyle g(x)=2x^2-16x+15 }\) \(\large\color{black}{ \displaystyle g(x)=2(x^2-8x)+15 }\) following so far?

  78. anonymous
    • one year ago
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    So i don't get g(x)

  79. SolomonZelman
    • one year ago
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    we will do g(x)... do you undertsand my previousppost?

  80. SolomonZelman
    • one year ago
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    previous post*

  81. anonymous
    • one year ago
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    yes

  82. SolomonZelman
    • one year ago
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    Ok. \(\large\color{black}{ \displaystyle g(x)=2(x^2-8x)+15 }\) what would you want to have added inside the parenthesis ?

  83. anonymous
    • one year ago
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    The x's?

  84. SolomonZelman
    • one year ago
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    what number do you need to add \(\large\color{black}{ \displaystyle g(x)=2(x^2-8x~+\color{blue}{\bf here} )+15 }\) to make the part in the parenthesis a perfect square trinomial ?

  85. anonymous
    • one year ago
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    2

  86. SolomonZelman
    • one year ago
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    when you have \(x^2-bx\) the number that you would like to add is \((b/2)^2\)

  87. SolomonZelman
    • one year ago
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    in this case (8/2)² --> 4² --> 16

  88. SolomonZelman
    • one year ago
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    So we would like to add 16, but we can't just add numbers, we are going to change the value of the function....

  89. SolomonZelman
    • one year ago
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    How would we add the number without changing the value of the function? like this....

  90. SolomonZelman
    • one year ago
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    \(\LARGE \color{black}{ \displaystyle g(x)=2(x^2-8x\color{blue}{+16}\color{red}{-16})+15 }\)

  91. SolomonZelman
    • one year ago
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    So far I haven't changed anything, I just added a "magic zero"

  92. anonymous
    • one year ago
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    ok I see

  93. SolomonZelman
    • one year ago
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    now, expand the -16 (in red) out of parenthesis \(\LARGE \color{black}{ \displaystyle g(x)=2(x^2-8x\color{blue}{+16})+2\cdot (\color{red}{-16})+15 }\) \(\LARGE \color{black}{ \displaystyle g(x)=2(x^2-8x\color{blue}{+16})-32+15 }\) \(\LARGE \color{black}{ \displaystyle g(x)=2(x^2-8x\color{blue}{+16})-17 }\) \(\LARGE \color{black}{ \displaystyle g(x)=2(x-4)^2-17 }\)

  94. SolomonZelman
    • one year ago
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    so what is the axis of summetry for g(x) ?

  95. anonymous
    • one year ago
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    4

  96. anonymous
    • one year ago
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    so would be f(x), hx, gx

  97. SolomonZelman
    • one year ago
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    yes, now you need to list all axis of summetry in order from smallest to greatest.

  98. SolomonZelman
    • one year ago
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    go ahead...

  99. anonymous
    • one year ago
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    fx, hx, gx

  100. anonymous
    • one year ago
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    yes? No? Maybe so?

  101. SolomonZelman
    • one year ago
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    yes. f , h, g. -4, 1, 4.

  102. anonymous
    • one year ago
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    Thank you so much:)

  103. SolomonZelman
    • one year ago
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    yw

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