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

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