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anonymous

  • one year ago

1. f(x) = 3 - x2 - 6x 2. (x) = x2 - 8x + 2 I have these two problems. I need to find out the domain, range, maximum, and minimum for both of the above. I keep getting the wrong answers when I try it, and some that don't make sense.. Please help? Thanks in advance!

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  1. anonymous
    • one year ago
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    Can someone help me please!!?

  2. SolomonZelman
    • one year ago
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    For a parabola with a positive leading coefficient, absolute maximum won't exit. (it opens up, and its absolute minimum is the vertex - y-coordinate is the value of the minimum, and x-coordintae is where this minimum is located) ---------------------------------------------------- For a parabola with a negative leading coefficient, absolute minimum won't exit. (it opens up, and its absolute maximum is the vertex - y-coordinate is the value of the maximum, and x-coordintae is where this maximum is located) ---------------------------------------------------- To find the vertex in each parabola, you need to complete the square. (for each function)

  3. SolomonZelman
    • one year ago
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    ``` If you are not familiar with what "perfect square trinomial" means, then I would advise to review that concept (here, with other people, or watch a video, read a book, get a tutor.... idk, that is your responsibility. I won't do it now, because I got to go pretty soon). ```

  4. jtvatsim
    • one year ago
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    @macky342 You still there? Have you made any progress on the question, or still stuck? :)

  5. anonymous
    • one year ago
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    I'm still stuck unfortunately.

  6. jtvatsim
    • one year ago
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    K, let's see if we can get somewhere...

  7. anonymous
    • one year ago
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    math has always been my worst subject. thank you.

  8. jtvatsim
    • one year ago
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    Well, maybe we'll be able to turn that around. :) I'm taking a look...

  9. anonymous
    • one year ago
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    Great! Thank you. This is the last question i have in this class before i graduate so i just want to be done!

  10. jtvatsim
    • one year ago
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    First thing I'm going to do is rewrite the equations this way: \[f(x) = -x^2-6x+3\] and \[g(x) = x^2-8x+2\].

  11. anonymous
    • one year ago
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    okay, now what do i do?

  12. jtvatsim
    • one year ago
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    Alright, personally, for me a picture is worth a thousand words, so I'm going to set these up to make it easier to graph a picture.

  13. jtvatsim
    • one year ago
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    This is a trick that most classes don't teach you, but I'm going to factor an x out of the first two terms of each. The result looks like this:

  14. jtvatsim
    • one year ago
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    \[f(x)=-x(x+6)+3\] and \[g(x)=x(x-8)+2\]

  15. jtvatsim
    • one year ago
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    You probably haven't seen that, but does that make sense so far?

  16. anonymous
    • one year ago
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    not really :(

  17. jtvatsim
    • one year ago
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    OK, no worries, all I did is this for the first one \[ f(x) = -x^2-6x+3 = x(-x-6)+3 = -x(x+6)+3 \] So, I just ignored the 3, and factored an x out of the first two terms. Then, I took out the negative sign (I didn't have to but it looks nicer). Please let me know if you have any questions on this part. We may need to review factoring. :)

  18. anonymous
    • one year ago
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    i am not understanding a thing thats going on. i reviewed the chapter over 5 times. had a tutor try and explain it and my last hope was this website and i feel like theres still no hope :(

  19. jtvatsim
    • one year ago
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    Gotcha. No worries, there's just a mental block somewhere that we have to sort out. I've had plenty before myself. :) Let's start more basic.

  20. anonymous
    • one year ago
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    perfect. thanks.

  21. jtvatsim
    • one year ago
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    I'm going to do a few diagnostic questions (totally different than the current question) to see where your understanding starts. Does this make sense? \[x^2 + x = x(x + 1)\]

  22. anonymous
    • one year ago
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    nope.

  23. jtvatsim
    • one year ago
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    OK, that's fine. Let's go simpler. :)

  24. anonymous
    • one year ago
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    the x's is where i get lost.

  25. anonymous
    • one year ago
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    I'm pretty good at basic math

  26. jtvatsim
    • one year ago
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    OK, so does this make more sense \[3^2 + 3 = 3(3+1)\]

  27. anonymous
    • one year ago
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    yes

  28. jtvatsim
    • one year ago
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    OK, so let's see if we can build on that understanding.... Let me think for a moment...

  29. Mertsj
    • one year ago
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    Do you know that these two equations have graphs that are parabolas?

  30. anonymous
    • one year ago
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    yes.

  31. Mertsj
    • one year ago
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    In the first one, the x^2 term has a negative coefficient so it will look like this:

  32. Mertsj
    • one year ago
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    |dw:1437101892911:dw|

  33. anonymous
    • one year ago
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    okay.

  34. Mertsj
    • one year ago
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    In the second one, the x^2 term is positive so it will look like this:

  35. Mertsj
    • one year ago
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    |dw:1437101949338:dw|

  36. Mertsj
    • one year ago
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    Does the first one have a high point or a low point?

  37. anonymous
    • one year ago
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    high

  38. Mertsj
    • one year ago
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    That high point is called a maximum point. It is also the vertex of the parabola.

  39. anonymous
    • one year ago
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    that makes sense.

  40. jtvatsim
    • one year ago
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    So, we know the first one has a maximum, but does it have a minimum too?

  41. anonymous
    • one year ago
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    when i was trying what i knew it didn't. and i didn't know if that was right or not.. theres a space to enter it on my school work but i don't know what to put there if theres no minimum.

  42. jtvatsim
    • one year ago
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    You are right. There is no minimum for the first graph. You can just put 'no minimum' or if you want to be fancy you can say that the graph's minimum goes to 'negative infinity'

  43. anonymous
    • one year ago
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    great! so what is the maximum? and i still need the range and the domain :(

  44. jtvatsim
    • one year ago
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    Good question. Let's see if we can track down the maximum first. Since you aren't comfortable with x's we are going to have to do this by "brute force" and plug in numbers until we see a pattern. :)

  45. anonymous
    • one year ago
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    oh gosh. okay.

  46. jtvatsim
    • one year ago
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    No worries, though let's just try some obvious numbers. I'm assuming you are comfortable with plugging in numbers for x? Like f(x) = -x^2 - 6x + 3 so if x = 0, then -0^2 - 6*0 + 3 = 3.

  47. jtvatsim
    • one year ago
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    Basically, I just transform the x's into whatever number I like. Does that make sense?

  48. anonymous
    • one year ago
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    kind of? I'm not too sure whats going on. I'm extremely tired right now. I'm a little older, i have a son who's a year and a half and he's sick so sleep and graduation are hard to balance on top of a child.

  49. anonymous
    • one year ago
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    I'm trying here. let me look again.

  50. jtvatsim
    • one year ago
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    Sure, I get the picture. Let me see if we can't get over the hump... How about this... You've heard the saying "X marks the spot" right?

  51. anonymous
    • one year ago
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    yes.

  52. jtvatsim
    • one year ago
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    In math, it's the same idea, "x" marks the spot where some number should be.

  53. anonymous
    • one year ago
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    yes yes..

  54. jtvatsim
    • one year ago
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    So, if I give you a formula, x + 10 you can replace the x (that marks the spot) with a number to see what I really mean.

  55. jtvatsim
    • one year ago
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    I could mean, 0 + 10, 1 + 10, 2 + 10, 3 + 10, or lots of other things. The point is, I can always turn a "x equation" into just simple math and numbers. I get to choose.

  56. anonymous
    • one year ago
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    im understanding so far.

  57. jtvatsim
    • one year ago
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    OK, so if we turn it up a notch, and I give you 5x + x now I have two x's that mark the spot. I can choose to replace these with a number, but it must be the same number. So, 5*1 + 1 or 5*2 + 2 but not, 5*1 + 50 or 5*3 + 100, I must pick the same number since the "x"s are the same.

  58. anonymous
    • one year ago
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    okay.

  59. jtvatsim
    • one year ago
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    Alright, so the formula we have has this symbol in it x^2. Do you know what this symbol means?

  60. anonymous
    • one year ago
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    yes the x has a exponent.

  61. jtvatsim
    • one year ago
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    Good, and exponents tell us to multiply that number that many times. For example, 2^2 = 2*2 5^2 = 5*5 and so on

  62. anonymous
    • one year ago
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    got it.

  63. jtvatsim
    • one year ago
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    OK, so let's see if we can figure out what on earth the original formula is saying. -x^2 - 6x + 3 we have two "x"s (that mark the spot) so we get to pick a number to put in both spots.

  64. anonymous
    • one year ago
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    ok, can we pick something easy like 2?

  65. jtvatsim
    • one year ago
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    Sure, let's do that to start.

  66. jtvatsim
    • one year ago
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    So we have: - 2^2 -6*2 + 3 and we need to figure out what this means.

  67. anonymous
    • one year ago
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    yes.

  68. jtvatsim
    • one year ago
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    Alright, order of operations Please Excuse My Dear Aunt Sally, parentheses, exponents, multiplication/division addition/subtraction

  69. jtvatsim
    • one year ago
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    So first, what is 2^2?

  70. anonymous
    • one year ago
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    4

  71. jtvatsim
    • one year ago
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    Good, so we have -4 - 6*2 + 3 next is multiplication

  72. anonymous
    • one year ago
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    ok so 6*2 right?

  73. jtvatsim
    • one year ago
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    that is right!

  74. anonymous
    • one year ago
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    so 12.

  75. jtvatsim
    • one year ago
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    Good! -4 -12 + 3 this is easy now

  76. jtvatsim
    • one year ago
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    "easier" that is... :) the negatives might be a little strange. :)

  77. anonymous
    • one year ago
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    ok will i miltiply -4 & -12 or what do i do since theres no symbol?

  78. anonymous
    • one year ago
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    multiply lol

  79. anonymous
    • one year ago
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    I'm decent when it comes to negatives.

  80. jtvatsim
    • one year ago
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    There's technically subtraction there.

  81. anonymous
    • one year ago
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    how?

  82. jtvatsim
    • one year ago
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    Remember it was -x^2 minus 6x plus 3 in the original formula -x^2 - 6x + 3

  83. anonymous
    • one year ago
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    ok got it

  84. jtvatsim
    • one year ago
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    So, we then get -4 - 12 + 3 = -16 + 3 = -13 I believe.

  85. jtvatsim
    • one year ago
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    This gets faster, don't worry. It takes a lot of words to describe the process. :)

  86. anonymous
    • one year ago
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    mhm...

  87. anonymous
    • one year ago
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    this is gonna take til 5am.

  88. jtvatsim
    • one year ago
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    Hopefully not. :) So, believe it or not we have actually found a small piece of the parabola, we took x = 2 and got -13 as the result.

  89. anonymous
    • one year ago
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    not even your fault, its mine..

  90. jtvatsim
    • one year ago
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    This is graphed on the xy-axis like this|dw:1437103634210:dw|

  91. anonymous
    • one year ago
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    yes

  92. jtvatsim
    • one year ago
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    OK, well we can keep plugging in numbers until the sun goes down (or comes up in your case), but I'm going to give you a small trick to use.

  93. anonymous
    • one year ago
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    great.

  94. jtvatsim
    • one year ago
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    Always begin by using x = 0. After that use x = "the number in front of the x in the formula". Here's what I mean for the second part.

  95. jtvatsim
    • one year ago
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    We have -x^2 - 6x + 3. We should use x = 0 and x = -6 because the -6 is sitting in front of the x without weird exponents.

  96. jtvatsim
    • one year ago
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    If we had x^2 + 10x + 3 we would pick x = 0 and x = 10. OK with that trick?

  97. anonymous
    • one year ago
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    yes

  98. jtvatsim
    • one year ago
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    OK, so let's get started plug in x = 0 into the formula -x^2 - 6x + 3 I will speed things up for you, you should get -0^2 - 6*0 + 3 = -0 - 0 + 3 = 3.

  99. jtvatsim
    • one year ago
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    OK with that?

  100. jtvatsim
    • one year ago
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    Actually scratch that, sorry. But I just found a better way that will take one step for you to use to find the maximum, we still need to plug in to use it though so our practice is not for nothing.

  101. anonymous
    • one year ago
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    ok

  102. jtvatsim
    • one year ago
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    The maximum or minimum of a parabola can be found by plugging in a special x. \[x = -\frac{b}{2a}\] where b is the number sitting in front of the x without an exponent, and a is the number in front the x with an exponent. So, in -x^2 - 6x + 3, a = -1 and b = -6. For x^2 + 10x + 40, a = 1 and b = 10. It's a bit of a magic trick why this works, but it works.

  103. anonymous
    • one year ago
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    now I'm lost

  104. anonymous
    • one year ago
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    i fell asleep there for a minute too so that isn't helping

  105. jtvatsim
    • one year ago
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    K, don't worry. I'm pulling out all the tricks I have here, Let me explain it one step at a time.

  106. anonymous
    • one year ago
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    ok

  107. jtvatsim
    • one year ago
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    We are looking at -x^2 - 6x + 3 right? What number is in front of the x^2? Well, there is -1 there because there is a negative sign and 1 is always in front of any number.

  108. anonymous
    • one year ago
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    yes

  109. jtvatsim
    • one year ago
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    And what number is in front of the "x"? We already said that this was -6.

  110. anonymous
    • one year ago
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    yup

  111. jtvatsim
    • one year ago
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    The magic recipe says we want an x that is equal to -b/2a. b is the number in front of the x a is the number in front of the x^2.

  112. anonymous
    • one year ago
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    yes

  113. jtvatsim
    • one year ago
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    So, we found that a = -1 (in front of x^2) and b = -6 (in front of x). The magic recipe is -b/2a = -(-6)/2(-1)

  114. jtvatsim
    • one year ago
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    This looks horrible, but if we just multiply the negatives in front we get -(-6)/2(-1) = 6/2(-1) and the 2 times -1 is -2 =6/-2 then we get = -3. This is the magic x.

  115. anonymous
    • one year ago
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    i kind of understand, lets just go with it.

  116. jtvatsim
    • one year ago
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    K, We now plug in x = -3 into the original formula -x^2 - 6x + 3. This will give us the maximum that we were looking for (for the last 5 hours) -(-3)^2 -6(-3) + 3 = -9 +18 + 3 = 9 + 3 = 12.

  117. anonymous
    • one year ago
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    I'm sorry. its been such a struggle with me. i know it.

  118. jtvatsim
    • one year ago
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    No, no, I'm sorry it's taken me so long to figure out the best way to help you. It's totally fine. :)

  119. anonymous
    • one year ago
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    i really appreciate it.. but now what?

  120. jtvatsim
    • one year ago
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    OK, well we celebrate that we finally have the maximum is is 12. We know that the graph has no minimum, so the range gets as large as 12 and as small as negative infinity. We write this as \[Range \ is \ (-\infty, 12]\]

  121. anonymous
    • one year ago
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    lol how do i do the infinity symbol?

  122. jtvatsim
    • one year ago
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    Is this an online assignment? maybe just type "-infinity" in words I'm sure the teacher will be fine with that (hopefully). :)

  123. anonymous
    • one year ago
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    yes. he will fix it. he's nice.

  124. anonymous
    • one year ago
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    what exactly is the domain?

  125. jtvatsim
    • one year ago
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    Cool. For the domain, this has to do with what types of numbers can you plug in for x. There are problems with formulas like 10 divided by x, because 10 divided by 0 doesn't make sense. However, with our formula, we are find and can plug in whatever number we want. The domain is "all real numbers" or (-infinity, +infinity)

  126. jtvatsim
    • one year ago
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    *fine not find. :)

  127. anonymous
    • one year ago
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    great!!

  128. anonymous
    • one year ago
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    the only problem is theres still one more equation lol

  129. jtvatsim
    • one year ago
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    Now, just for that last one... It's the same thing, just with different numbers. Let's slay this dragon once and for all. >:)

  130. anonymous
    • one year ago
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    yay!

  131. jtvatsim
    • one year ago
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    Alright, so now, x^2 - 8x + 2. What number in front of x^2? That's a = 1. What number in front of x? That's b = -8.

  132. anonymous
    • one year ago
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    yup

  133. jtvatsim
    • one year ago
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    What is the magic recipe? -b/2a That gives us -(-8)/2(1) = 8/2 = 4. The magic x is 4. (Remember that this second graph has a minimum? This magic recipe will always find the maximum OR the minimum and it knows which one is which! Crazy, but true).

  134. anonymous
    • one year ago
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    but doesn't the graph have no maximum?

  135. jtvatsim
    • one year ago
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    That's right. So this magic x will give us the minimum that we want. Pretty cool of it. The reason this works is because of Calculus, but let's not go there today... :)

  136. anonymous
    • one year ago
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    thank god i don't need anymore math to graduate. lol

  137. jtvatsim
    • one year ago
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    lol! Anyways, we now take x = 4, and replace this in our original formula x^2 - 8x + 2 4^2 - 8(4) + 2 = 16 - 32 + 2 = -16 + 2 = -14 if I'm not mistaken.

  138. anonymous
    • one year ago
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    uh huh uh huh

  139. jtvatsim
    • one year ago
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    lol

  140. anonymous
    • one year ago
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    -14 is the minimum correct?

  141. jtvatsim
    • one year ago
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    Yep! So, the minimum is -14. There is no maximum (or its +infinity). The range is [-14, +infinity). The domain is fine so "all real numbers" or (-infinity,+infinity).

  142. jtvatsim
    • one year ago
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    I'm going to graph this on my graphing calculator just to confirm. :)

  143. anonymous
    • one year ago
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    great!! i really appreciate all of your help!!

  144. jtvatsim
    • one year ago
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    You are very welcome! I'm glad the dragon is dead now. lol :) Here's the second graph, and we are right.

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  145. anonymous
    • one year ago
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    yay!!

  146. jtvatsim
    • one year ago
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    and the first one, is...... right again! Yay!

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  147. anonymous
    • one year ago
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    great your amazing!! i appreciate it so much!! especially the patience you had!!

  148. jtvatsim
    • one year ago
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    No problem! I'm glad to help! Great effort on your end as well, you stuck it out and survived to tell the tale. :)

  149. jtvatsim
    • one year ago
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    Go get some rest, your brain is probably deep fried. :)

  150. anonymous
    • one year ago
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    Thanks again! :D

  151. jtvatsim
    • one year ago
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    Take care! :D

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