anonymous
  • anonymous
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!
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Can someone help me please!!?
SolomonZelman
  • 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)
SolomonZelman
  • 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). ```

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jtvatsim
  • jtvatsim
@macky342 You still there? Have you made any progress on the question, or still stuck? :)
anonymous
  • anonymous
I'm still stuck unfortunately.
jtvatsim
  • jtvatsim
K, let's see if we can get somewhere...
anonymous
  • anonymous
math has always been my worst subject. thank you.
jtvatsim
  • jtvatsim
Well, maybe we'll be able to turn that around. :) I'm taking a look...
anonymous
  • anonymous
Great! Thank you. This is the last question i have in this class before i graduate so i just want to be done!
jtvatsim
  • jtvatsim
First thing I'm going to do is rewrite the equations this way: \[f(x) = -x^2-6x+3\] and \[g(x) = x^2-8x+2\].
anonymous
  • anonymous
okay, now what do i do?
jtvatsim
  • 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.
jtvatsim
  • 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:
jtvatsim
  • jtvatsim
\[f(x)=-x(x+6)+3\] and \[g(x)=x(x-8)+2\]
jtvatsim
  • jtvatsim
You probably haven't seen that, but does that make sense so far?
anonymous
  • anonymous
not really :(
jtvatsim
  • 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. :)
anonymous
  • 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 :(
jtvatsim
  • 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.
anonymous
  • anonymous
perfect. thanks.
jtvatsim
  • 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? \[x^2 + x = x(x + 1)\]
anonymous
  • anonymous
nope.
jtvatsim
  • jtvatsim
OK, that's fine. Let's go simpler. :)
anonymous
  • anonymous
the x's is where i get lost.
anonymous
  • anonymous
I'm pretty good at basic math
jtvatsim
  • jtvatsim
OK, so does this make more sense \[3^2 + 3 = 3(3+1)\]
anonymous
  • anonymous
yes
jtvatsim
  • jtvatsim
OK, so let's see if we can build on that understanding.... Let me think for a moment...
Mertsj
  • Mertsj
Do you know that these two equations have graphs that are parabolas?
anonymous
  • anonymous
yes.
Mertsj
  • Mertsj
In the first one, the x^2 term has a negative coefficient so it will look like this:
Mertsj
  • Mertsj
|dw:1437101892911:dw|
anonymous
  • anonymous
okay.
Mertsj
  • Mertsj
In the second one, the x^2 term is positive so it will look like this:
Mertsj
  • Mertsj
|dw:1437101949338:dw|
Mertsj
  • Mertsj
Does the first one have a high point or a low point?
anonymous
  • anonymous
high
Mertsj
  • Mertsj
That high point is called a maximum point. It is also the vertex of the parabola.
anonymous
  • anonymous
that makes sense.
jtvatsim
  • jtvatsim
So, we know the first one has a maximum, but does it have a minimum too?
anonymous
  • 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.
jtvatsim
  • 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'
anonymous
  • anonymous
great! so what is the maximum? and i still need the range and the domain :(
jtvatsim
  • 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. :)
anonymous
  • anonymous
oh gosh. okay.
jtvatsim
  • 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.
jtvatsim
  • jtvatsim
Basically, I just transform the x's into whatever number I like. Does that make sense?
anonymous
  • 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.
anonymous
  • anonymous
I'm trying here. let me look again.
jtvatsim
  • 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?
anonymous
  • anonymous
yes.
jtvatsim
  • jtvatsim
In math, it's the same idea, "x" marks the spot where some number should be.
anonymous
  • anonymous
yes yes..
jtvatsim
  • 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.
jtvatsim
  • 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.
anonymous
  • anonymous
im understanding so far.
jtvatsim
  • 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.
anonymous
  • anonymous
okay.
jtvatsim
  • jtvatsim
Alright, so the formula we have has this symbol in it x^2. Do you know what this symbol means?
anonymous
  • anonymous
yes the x has a exponent.
jtvatsim
  • 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
anonymous
  • anonymous
got it.
jtvatsim
  • 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.
anonymous
  • anonymous
ok, can we pick something easy like 2?
jtvatsim
  • jtvatsim
Sure, let's do that to start.
jtvatsim
  • jtvatsim
So we have: - 2^2 -6*2 + 3 and we need to figure out what this means.
anonymous
  • anonymous
yes.
jtvatsim
  • jtvatsim
Alright, order of operations Please Excuse My Dear Aunt Sally, parentheses, exponents, multiplication/division addition/subtraction
jtvatsim
  • jtvatsim
So first, what is 2^2?
anonymous
  • anonymous
4
jtvatsim
  • jtvatsim
Good, so we have -4 - 6*2 + 3 next is multiplication
anonymous
  • anonymous
ok so 6*2 right?
jtvatsim
  • jtvatsim
that is right!
anonymous
  • anonymous
so 12.
jtvatsim
  • jtvatsim
Good! -4 -12 + 3 this is easy now
jtvatsim
  • jtvatsim
"easier" that is... :) the negatives might be a little strange. :)
anonymous
  • anonymous
ok will i miltiply -4 & -12 or what do i do since theres no symbol?
anonymous
  • anonymous
multiply lol
anonymous
  • anonymous
I'm decent when it comes to negatives.
jtvatsim
  • jtvatsim
There's technically subtraction there.
anonymous
  • anonymous
how?
jtvatsim
  • jtvatsim
Remember it was -x^2 minus 6x plus 3 in the original formula -x^2 - 6x + 3
anonymous
  • anonymous
ok got it
jtvatsim
  • jtvatsim
So, we then get -4 - 12 + 3 = -16 + 3 = -13 I believe.
jtvatsim
  • jtvatsim
This gets faster, don't worry. It takes a lot of words to describe the process. :)
anonymous
  • anonymous
mhm...
anonymous
  • anonymous
this is gonna take til 5am.
jtvatsim
  • 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.
anonymous
  • anonymous
not even your fault, its mine..
jtvatsim
  • jtvatsim
This is graphed on the xy-axis like this|dw:1437103634210:dw|
anonymous
  • anonymous
yes
jtvatsim
  • 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.
anonymous
  • anonymous
great.
jtvatsim
  • 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.
jtvatsim
  • 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.
jtvatsim
  • jtvatsim
If we had x^2 + 10x + 3 we would pick x = 0 and x = 10. OK with that trick?
anonymous
  • anonymous
yes
jtvatsim
  • 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.
jtvatsim
  • jtvatsim
OK with that?
jtvatsim
  • 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.
anonymous
  • anonymous
ok
jtvatsim
  • 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.
anonymous
  • anonymous
now I'm lost
anonymous
  • anonymous
i fell asleep there for a minute too so that isn't helping
jtvatsim
  • jtvatsim
K, don't worry. I'm pulling out all the tricks I have here, Let me explain it one step at a time.
anonymous
  • anonymous
ok
jtvatsim
  • 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.
anonymous
  • anonymous
yes
jtvatsim
  • jtvatsim
And what number is in front of the "x"? We already said that this was -6.
anonymous
  • anonymous
yup
jtvatsim
  • 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.
anonymous
  • anonymous
yes
jtvatsim
  • 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)
jtvatsim
  • 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.
anonymous
  • anonymous
i kind of understand, lets just go with it.
jtvatsim
  • 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.
anonymous
  • anonymous
I'm sorry. its been such a struggle with me. i know it.
jtvatsim
  • 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. :)
anonymous
  • anonymous
i really appreciate it.. but now what?
jtvatsim
  • 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 \[Range \ is \ (-\infty, 12]\]
anonymous
  • anonymous
lol how do i do the infinity symbol?
jtvatsim
  • jtvatsim
Is this an online assignment? maybe just type "-infinity" in words I'm sure the teacher will be fine with that (hopefully). :)
anonymous
  • anonymous
yes. he will fix it. he's nice.
anonymous
  • anonymous
what exactly is the domain?
jtvatsim
  • 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)
jtvatsim
  • jtvatsim
*fine not find. :)
anonymous
  • anonymous
great!!
anonymous
  • anonymous
the only problem is theres still one more equation lol
jtvatsim
  • 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. >:)
anonymous
  • anonymous
yay!
jtvatsim
  • 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.
anonymous
  • anonymous
yup
jtvatsim
  • 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).
anonymous
  • anonymous
but doesn't the graph have no maximum?
jtvatsim
  • 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... :)
anonymous
  • anonymous
thank god i don't need anymore math to graduate. lol
jtvatsim
  • 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.
anonymous
  • anonymous
uh huh uh huh
jtvatsim
  • jtvatsim
lol
anonymous
  • anonymous
-14 is the minimum correct?
jtvatsim
  • 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).
jtvatsim
  • jtvatsim
I'm going to graph this on my graphing calculator just to confirm. :)
anonymous
  • anonymous
great!! i really appreciate all of your help!!
jtvatsim
  • jtvatsim
You are very welcome! I'm glad the dragon is dead now. lol :) Here's the second graph, and we are right.
1 Attachment
anonymous
  • anonymous
yay!!
jtvatsim
  • jtvatsim
and the first one, is...... right again! Yay!
1 Attachment
anonymous
  • anonymous
great your amazing!! i appreciate it so much!! especially the patience you had!!
jtvatsim
  • 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. :)
jtvatsim
  • jtvatsim
Go get some rest, your brain is probably deep fried. :)
anonymous
  • anonymous
Thanks again! :D
jtvatsim
  • jtvatsim
Take care! :D

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