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Help MasterA with Math homework ;)
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| MasterA |
Posted on 10/03/2011 18:35:21
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Super Admin
Posts: 2582
Joined: 21.07.11
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hello everyone ive been running behind in my homework recently (beta server ftw) and ive ran across a few problems ive been having trouble on and knowing all you math pros here maybe you can help me <3  . its due tommorrow so i need help asap! well here they are any help even on a few will be much appreciated 
1.A parabola has a line of symmetry at x=3 and contains the points (1,0) and (4,-3). What is the function for this graph.
2.An antique is worth $1000 after 2 years and is predicted to be worth $1300 in 5 years. What is its predicted value in 8 years.
3.what is the remainder of ((x^29) -(7x^14) +8)/(x-1)
4.write the complex number in standard form (1-i)^3
5.write the polynomial funcion of minimum degree in standard form with real coefficients whose zeros ad thier multiplicity include those listed
1(multiplicity 2), -2(multiplicity 3)
6.state how many complex and real zeros the function has
f(x)=x^4 -2x^2 +3x -4
7.find all the zeros and write a linear factorization of the function
f(x)=3x^4 +8x^3 +6x^2+3x-2
8.true or false: A polynomial degree 3 with real coefficients must have two nonreal zeros. Justify your answer.
Edited by MasterA on 10/03/2011 18:45:24 |
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| ZA numpty |
Posted on 10/03/2011 18:40:49
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ShawnPeezy
Posts: 4430
Joined: 16.11.10
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... OW MY HEAD OW OW OW OW OW OW |
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| nygiants88 |
Posted on 10/03/2011 19:04:46
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Super Admin
Posts: 1707
Joined: 24.07.11
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6.state how many complex and real zeros the function has
f(x)=x^4 -2x^2 +3x -4
Its a trick question! there are no zeros.
If you take the literal interpretation of this problem the answer is zero.
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| MasterA |
Posted on 10/03/2011 19:07:40
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Posts: 2582
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nygiants88 wrote:
6.state how many complex and real zeros the function has
f(x)=x^4 -2x^2 +3x -4
Its a trick question! there are no zeros.
If you take the literal interpretation of this problem the answer is zero.
It does have zeros because it passes through the x-axis its just the whole imaginary i zeros i have to find with it which is tough :/ |
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| ZA RedXlll |
Posted on 10/03/2011 19:11:56
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Numpty
Posts: 1181
Joined: 30.12.10
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I would help if i could but i havent taken a legit algebra class in about 2 years. Now if this was geometry i could do this in a heartbeat. But i suggest you use a graphing calculator. That was my cheat for most of that class xD
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Last Updated: 11/14/11 |
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| MasterA |
Posted on 10/03/2011 19:17:36
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Posts: 2582
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haha thanks red i do use a graphing clac by the way just some of these problems are killing me ugh no halo tonight i suppose lol |
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| ZA Bvigil |
Posted on 10/03/2011 19:46:15
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Admin
Posts: 2161
Joined: 15.11.10
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I can help with all of these if you still need it
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| MasterA |
Posted on 10/03/2011 19:49:07
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Posts: 2582
Joined: 21.07.11
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that would be much i appericated i still can't figure them out |
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| alltheabove |
Posted on 10/03/2011 19:50:42
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Senior Member
Posts: 410
Joined: 20.06.11
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I actually don't have much homework tonight, ftw! So...
*DISCLAIMER* I don't guarantee that my answer are correct >_>
1. Given:
Line of symmetry @ x=3
points (1,0) and (4, -3)
That being said, the line of symmetry's equation is:
x= -b/(2a)
therefore, 3= -b/(2a)
Then, you take the standard form of a parabola:
ax^2 + bx + c = 0
Plug in the two given points:
a(1)^2 + 1b + c = 0 for (1,0)
a(4)^2 + 4b + c = -3 for (4, -3)
which simplifies to be:
a + b + c = 0
16a + 4b + c = -3
Now, you have a system of equations!
16a + 4b + c = -3
a + b + c = 0
Subtract the two to get 15a + 3b = -3.
Now back to the line of symmetry; 3= -b/(2a)
Therefore, 6a = -b, so b = -6a
Plug that in, and you get
15a - 18a = -3
Simplified:
-3a = -3
So, a = 1.
b = -6a, so b = -6.
a + b + c = 0, so:
1 - 6 + c = 0 ---> -5 + c = 0 ---> c = 5.
YAY, a=1, b=-6, c=5.
Final result, x^2 - 6x + 5 = 0.
Check my work, it might be wrong. Also, sorry for so much spacing. Imma see if I can do others later.
*EDIT* Checked with the two given points, it seems right.
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| MasterA |
Posted on 10/03/2011 19:54:52
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Super Admin
Posts: 2582
Joined: 21.07.11
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alltheabove wrote:
I actually don't have much homework tonight, ftw! So...
*DISCLAIMER* I don't guarantee that my answer are correct >_>
1. Given:
Line of symmetry @ x=3
points (1,0) and (4, -3)
That being said, the line of symmetry's equation is:
x= -b/(2a)
therefore, 3= -b/(2a)
Then, you take the standard form of a parabola:
ax^2 + bx + c = 0
Plug in the two given points:
a(1)^2 + 1b + c = 0 for (1,0)
a(4)^2 + 4b + c = -3 for (4, -3)
which simplifies to be:
a + b + c = 0
16a + 4b + c = -3
Now, you have a system of equations!
16a + 4b + c = -3
a + b + c = 0
Subtract the two to get 15a + 3b = -3.
Now back to the line of symmetry; 3= -b/(2a)
Therefore, 6a = -b, so b = -6a
Plug that in, and you get
15a - 18a = -3
Simplified:
-3a = -3
So, a = 1.
b = -6a, so b = -6.
a + b + c = 0, so:
1 - 6 + c = 0 ---> -5 + c = 0 ---> c = 5.
YAY, a=1, b=-6, c=5.
Final result, x^2 - 6x + 5 = 0.
Check my work, it might be wrong. Also, sorry for so much spacing. Imma see if I can do others later.
*EDIT* Checked with the two given points, it seems right.
i plugged it in and graphed it and it met all the requirements thanks for explaining it with along with the answer above your a great help |
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| ZA Bvigil |
Posted on 10/03/2011 19:57:03
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Admin
Posts: 2161
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WARNING WARNING I WILL BE SPAMMING THIS THREAD AS I POST HOW TO DO THESE PROBLEMS
1. First thing you need to do is find the zeros which are located at 1 and 5 because the line of symmetry is at 3
2. Now you have the equation of y=(x-1)(x-5)
now you can either leave it there or FOIL it out and get y=x^2-6x+5 There you go
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| alltheabove |
Posted on 10/03/2011 19:58:35
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Senior Member
Posts: 410
Joined: 20.06.11
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Number two's straightforward. Imma make a new post cuz I dislike cluttering >__>
2. Given:
Worth $1000 after 2 years
Worth $1300 in 5 years
What this essentially means is that you have two points. Since time is almost always graphed on the x-axis, x=time. Therefore, the two given points are:
(2, 1000)
(5, 1300)
Find the slope!
(1300-1000)
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(5-2)
= 300/3 = 100.
Typical y=mx+b, so right now:
y = 100x + b
Finding b's easy, just plug in a point. I'll use (2, 1000):
1000 = 100(2) + b ---> 1000 = 200 + b ---> b = 800
Equation's done! y = 100x + 800
Now, since time = x, and you're given that time = 8 years...
y = 100(8) + 800
wewt answer, y = 800 + 800, so $1600.
This is assuming that the relationship is linear. If it's anything but linear, I probably wouldn't be able to do it without a calculator...
Once more, check my work.
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| ZA Bvigil |
Posted on 10/03/2011 20:04:54
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Admin
Posts: 2161
Joined: 15.11.10
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Both Alltheaboves and my system work but mine is faster (:
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| MasterA |
Posted on 10/03/2011 20:06:42
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Super Admin
Posts: 2582
Joined: 21.07.11
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thanks too vigil im really glad for the help |
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| alltheabove |
Posted on 10/03/2011 20:11:51
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Senior Member
Posts: 410
Joined: 20.06.11
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Can you check the prompt for number 3? If it really is as you posted, there's gotta be an easier way to do it, else long division's gonna be one pain in the ass.
So for now, number 4. Please note that root(#) means square root.
First, "i" has a cyclical nature, meaning:
i = root(-1)
i^2 = -1
i^3 = -i = -root(-1)
i^4 = 1
i^5 = root(-1)
and so forth. If you multiply root(-1) for yourself, you'll see that this is pretty straightforward.
That being said, the problem becomes easyyy! I'll leave everything in terms of "i" until the end xD
(1-i)^3 ---> (1-i) x (1-i) x (1-i)
soooo, first step would be to multiply the first two, getting:
(1-i) x (1 - i - i + i^2) ---> (1-i) x (1 - 2i + i^2)
Keep going (foil it!), and you get:
1 - 2i + i^2 - i + 3i - i^3
Note that you have a -2i, -i, and a +3i, which cancel out. Yay.
Now, you have:
1 + i^2 - i^3
Plug in the table above, and you get:
1 + (-1) - (-root(-1))
So you just end up with -(-i^3), or +root(-1).
Check it, I'm a careless person in math D:
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| MasterA |
Posted on 10/03/2011 20:19:54
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Super Admin
Posts: 2582
Joined: 21.07.11
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soooo, first step would be to multiply the first two, getting:
(1-i) x (1 - i - i + i^2) ---> (1-i) x (1 - 2i + i^2)
Keep going (foil it!), and you get:
1 - 2i + i^2 - i + 3i - i^3
when you foil it would the -i x the -2i be 3i^2? |
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| MasterA |
Posted on 10/03/2011 20:23:33
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Super Admin
Posts: 2582
Joined: 21.07.11
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oh and about number 3 thats it my friend solved it by using synthetic division using the first number then alot of zeros and doing the same thing for the ^14 but i didnt think that was the right way to approach it :/ |
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| alltheabove |
Posted on 10/03/2011 20:31:12
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Senior Member
Posts: 410
Joined: 20.06.11
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Number 5's simple, just a little tedious with a lot of room for careless errors.
And I had to look up with multiplicity meant lol xD
So...
Given: problem as stated above, lazy me.
Two things are critical in doing this problem:
1. Multiplicity- in case you didn't know, that's just how many times a zero occurs. Ex. a zero of x=1 with a multiplicity of 2 means the graph has a recurring zero at x=1, twice.
2. Be sure to remember that the sign flips when putting the zero into form. Ex. a zero of x=1 means (x-1), NOT (x+1).
That being said, the basic equation is ((x-1)^2) x ((x+2)^3). Please note that the x in the middle denotes multiply O:
If that's standard form for your teacher, woohoo! If not, expansion time...This is the way that I like to do it. Dunno how you do:
First, I expand out the entire thing:
(x-1) (x-1) (x+2) (x+2) (x+2)
Then, I take two of the like terms (x-1) and multiply them. I'll assume that you can combine like terms to avoid unnecessary space.
(x^2 - 2x + 1) (x + 2) (x + 2) (x + 2)
Now two more like terms (x+2) for the following:
(x^2 - 2x + 1) (x^2 + 4x + 4) (x + 2)
Then, I like to combine what's left of the like terms, meaning the (x+2) with the (x+2)^2, begetting:
(x^2 - 2x + 1) (x^3 + 4x^2 + 4x + 4x^2 + 8x + 8)
Simplify. Whenever I do this, I like to circle the exponent that I'm working on (including the sign), and cross the terms out after I'm done with them:
(x^2 - 2x + 1) (x^3 + 8x^2 + 12x + 8)
Whee, fun time...foil it!
x^5 + 8x^4 + 12x^3 + 8x^2 - 2x^4 - 8x^3 - 24x^2 - 16x + x^3 + 8x^2 + 12x + 8
Simplify...........:
x^5 + 6x^4 + 5x^3 - 24x^2 - 4x + 8
If ever there was a problem to double check, it would be this one. O:
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| alltheabove |
Posted on 10/03/2011 20:34:43
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Senior Member
Posts: 410
Joined: 20.06.11
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MasterA wrote:
soooo, first step would be to multiply the first two, getting:
(1-i) x (1 - i - i + i^2) ---> (1-i) x (1 - 2i + i^2)
Keep going (foil it!), and you get:
1 - 2i + i^2 - i + 3i - i^3
when you foil it would the -i x the -2i be 3i^2?
Seeeee I knew I would mess up fml.
Oh well, foil it out, and you should ACTUALLY get:
1 - 2i + i^2 - i + 2i^2 - i^3
Therefore, answer's not so simple T___T Simplified:
1 - 3i + 3i^2 - i^3
Then plug in table values for standard form!
*EDIT* Yeah, you could use synthetic division for number three, using the root and a whole buncha 0's. I was looking for another way, but I don't think there is one. This means that you'd have x^29, then a whole bunch of zeros until you hit x^14, then a whole buncha zeros until you finish out the rest of the number.
Edited by alltheabove on 10/03/2011 20:46:29
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| MasterA |
Posted on 10/03/2011 20:46:10
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Super Admin
Posts: 2582
Joined: 21.07.11
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alright thanks man your really good at this lol i have like 8 sections with 15 problems apeice that ive been doing for the past 4 and a half hours lol thanks for helping with these hard ones |
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