## 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!

1. anonymous

2. SolomonZelman

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

 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

@macky342 You still there? Have you made any progress on the question, or still stuck? :)

5. anonymous

I'm still stuck unfortunately.

6. jtvatsim

K, let's see if we can get somewhere...

7. anonymous

math has always been my worst subject. thank you.

8. jtvatsim

Well, maybe we'll be able to turn that around. :) I'm taking a look...

9. anonymous

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

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

okay, now what do i do?

12. jtvatsim

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

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

$f(x)=-x(x+6)+3$ and $g(x)=x(x-8)+2$

15. jtvatsim

You probably haven't seen that, but does that make sense so far?

16. anonymous

not really :(

17. jtvatsim

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

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

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

perfect. thanks.

21. jtvatsim

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

nope.

23. jtvatsim

OK, that's fine. Let's go simpler. :)

24. anonymous

the x's is where i get lost.

25. anonymous

I'm pretty good at basic math

26. jtvatsim

OK, so does this make more sense $3^2 + 3 = 3(3+1)$

27. anonymous

yes

28. jtvatsim

OK, so let's see if we can build on that understanding.... Let me think for a moment...

29. Mertsj

Do you know that these two equations have graphs that are parabolas?

30. anonymous

yes.

31. Mertsj

In the first one, the x^2 term has a negative coefficient so it will look like this:

32. Mertsj

|dw:1437101892911:dw|

33. anonymous

okay.

34. Mertsj

In the second one, the x^2 term is positive so it will look like this:

35. Mertsj

|dw:1437101949338:dw|

36. Mertsj

Does the first one have a high point or a low point?

37. anonymous

high

38. Mertsj

That high point is called a maximum point. It is also the vertex of the parabola.

39. anonymous

that makes sense.

40. jtvatsim

So, we know the first one has a maximum, but does it have a minimum too?

41. anonymous

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

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

great! so what is the maximum? and i still need the range and the domain :(

44. jtvatsim

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

oh gosh. okay.

46. jtvatsim

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

Basically, I just transform the x's into whatever number I like. Does that make sense?

48. anonymous

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

I'm trying here. let me look again.

50. jtvatsim

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

yes.

52. jtvatsim

In math, it's the same idea, "x" marks the spot where some number should be.

53. anonymous

yes yes..

54. jtvatsim

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

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

im understanding so far.

57. jtvatsim

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

okay.

59. jtvatsim

Alright, so the formula we have has this symbol in it x^2. Do you know what this symbol means?

60. anonymous

yes the x has a exponent.

61. jtvatsim

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

got it.

63. jtvatsim

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

ok, can we pick something easy like 2?

65. jtvatsim

Sure, let's do that to start.

66. jtvatsim

So we have: - 2^2 -6*2 + 3 and we need to figure out what this means.

67. anonymous

yes.

68. jtvatsim

Alright, order of operations Please Excuse My Dear Aunt Sally, parentheses, exponents, multiplication/division addition/subtraction

69. jtvatsim

So first, what is 2^2?

70. anonymous

4

71. jtvatsim

Good, so we have -4 - 6*2 + 3 next is multiplication

72. anonymous

ok so 6*2 right?

73. jtvatsim

that is right!

74. anonymous

so 12.

75. jtvatsim

Good! -4 -12 + 3 this is easy now

76. jtvatsim

"easier" that is... :) the negatives might be a little strange. :)

77. anonymous

ok will i miltiply -4 & -12 or what do i do since theres no symbol?

78. anonymous

multiply lol

79. anonymous

I'm decent when it comes to negatives.

80. jtvatsim

There's technically subtraction there.

81. anonymous

how?

82. jtvatsim

Remember it was -x^2 minus 6x plus 3 in the original formula -x^2 - 6x + 3

83. anonymous

ok got it

84. jtvatsim

So, we then get -4 - 12 + 3 = -16 + 3 = -13 I believe.

85. jtvatsim

This gets faster, don't worry. It takes a lot of words to describe the process. :)

86. anonymous

mhm...

87. anonymous

this is gonna take til 5am.

88. jtvatsim

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

not even your fault, its mine..

90. jtvatsim

This is graphed on the xy-axis like this|dw:1437103634210:dw|

91. anonymous

yes

92. jtvatsim

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

great.

94. jtvatsim

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

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

If we had x^2 + 10x + 3 we would pick x = 0 and x = 10. OK with that trick?

97. anonymous

yes

98. jtvatsim

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

OK with that?

100. jtvatsim

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

ok

102. jtvatsim

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

now I'm lost

104. anonymous

i fell asleep there for a minute too so that isn't helping

105. jtvatsim

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

ok

107. jtvatsim

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

yes

109. jtvatsim

And what number is in front of the "x"? We already said that this was -6.

110. anonymous

yup

111. jtvatsim

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

yes

113. jtvatsim

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

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

i kind of understand, lets just go with it.

116. jtvatsim

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

I'm sorry. its been such a struggle with me. i know it.

118. jtvatsim

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

i really appreciate it.. but now what?

120. jtvatsim

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

lol how do i do the infinity symbol?

122. jtvatsim

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

123. anonymous

yes. he will fix it. he's nice.

124. anonymous

what exactly is the domain?

125. jtvatsim

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

*fine not find. :)

127. anonymous

great!!

128. anonymous

the only problem is theres still one more equation lol

129. jtvatsim

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

yay!

131. jtvatsim

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

yup

133. jtvatsim

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

but doesn't the graph have no maximum?

135. jtvatsim

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

thank god i don't need anymore math to graduate. lol

137. jtvatsim

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

uh huh uh huh

139. jtvatsim

lol

140. anonymous

-14 is the minimum correct?

141. jtvatsim

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

I'm going to graph this on my graphing calculator just to confirm. :)

143. anonymous

great!! i really appreciate all of your help!!

144. jtvatsim

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

145. anonymous

yay!!

146. jtvatsim

and the first one, is...... right again! Yay!

147. anonymous

great your amazing!! i appreciate it so much!! especially the patience you had!!

148. jtvatsim

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

Go get some rest, your brain is probably deep fried. :)

150. anonymous

Thanks again! :D

151. jtvatsim

Take care! :D