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garrett01001
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Joined: 21 Apr 2007
Posts: 135
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 Posted: 06 May 2007 02:24:54 pm Post subject: |
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my teacher gave me this on my homework:
"Write the equation of a circle in standard form and give the center and radius.
4x²+4y²-24x+8y+39=0. State the domain and range."
that is exactly how it is written on my paper...
she has taught us circles, but nothing that looks like this.. and she refused to help me.. I can't even find anything like this in the book that she gave us..
any ideas on how to solve this without using square roots? or even how to solve it with square roots?
I'm not asking for the answer, just give me some hints as to what the heck I'm supposed to do.. |
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frenchcalc1
جان ألعريم
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Joined: 14 Mar 2007
Posts: 648
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| Posted: 06 May 2007 02:29:45 pm Post subject: |
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You do know that the standard equation of a circle is [font="courier new;font-size:9pt;line-height:100%;color:darkblue"](x-h)² + (y-k)² = r² right? And you do know that [font="courier new;font-size:9pt;line-height:100%;color:darkblue"](h,k) is the center of the circle, and that [font="courier new;font-size:9pt;line-height:100%;color:darkblue"]r² is the radius squared? If so, then in order to get to this equation, all you have to do is factor the general form(which is what your teacher gave you) into standard ;)
Quote: Answer:
(x-3)² + (y+1)² = 1/4
Last edited by Guest on 06 May 2007 02:41:49 pm; edited 1 time in total |
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garrett01001
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Joined: 21 Apr 2007
Posts: 135
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| Posted: 07 May 2007 12:07:12 am Post subject: |
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err.. how did you get = 1/4 ??
I got: (x-3)²+(y+1)²=17/4
Steps:
4x²+4y²-24x+8y+39=0
4x²-24x+4y²+8y+39=0
(4x²-24x+(-24/4)²)+(4y²+8y+(8/4)²)+39=(-24/4)²+(8/4)²
(4x²-24x+(-24/4)²)+(4y²+8y+(8/4)²)=40-39
(4x²+(-6)²)+(4y²+2²)=1+24-8
(2x-6)²+(2y+2)²=17
(x-3)²+(y+1)²=17/4
not sure if all of these steps are in the right order.. but I got the answer.. i hope..
maybe you just typo'd?
(note: did this before seeing you had given the answer..)
had to look up general form for circles on the net to even figure out wth needed to be done.. my book definately does not even mention this.. (i stay 2 chapters ahead of the class..)
Im going to have to slap my teacher with a TI for that one.. (not mine of course, I love her too much.. maybe one of the crappy 82's that she has for student use)
And, did you typo/intentionally put the wrong answer..?
Here's another problem from the same HW.. took me seven tries to get it right.. (i forgot to carry the square..)
"Prove that cos²2x-cos²x=sin²x-sin²2x"
my steps:
cos²2x-cos²x=sin²x-4sin²xcos²x
cos²2x=1-4sin²xcos²x
cos²2x=1-4sin²x(1-sin²x)
cos²2x=1-4sin²x+4sin^4x (note: cant find superscript of 4 and this forum doesnt have superscript tags.. 4sin^4x = 4(sinx)^4)
1-sin²x=1-4sin²x+4sin^4x
1-4sin²xcos²x= ...
1-4sin²x(1-sin²x)= ...
1-4sin²x+4sin^4x = 1-4sin²x+4sin^4x
very annoying problem 
Last edited by Guest on 07 May 2007 12:57:10 am; edited 1 time in total |
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thornahawk
μολών λαβέ
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Joined: 27 Mar 2005
Posts: 569
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| Posted: 07 May 2007 04:37:16 am Post subject: |
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Did your teacher not teach you "completing the square"? ;)
Anyway... the best way to check if the standard form you got is correct is to expand it back and compare with the original equation. ;)
About superscripts and subscripts... the sup and sub BBCode tags were made for that purpose. :)
thornahawk |
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frenchcalc1
جان ألعريم
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Joined: 14 Mar 2007
Posts: 648
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| Posted: 07 May 2007 11:19:58 am Post subject: |
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I keep getting 1/4...did I do something wrong? Algebra 2 (secondd semester) has rid me of my first semester calculating skills  |
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garrett01001
Member
Joined: 21 Apr 2007
Posts: 135
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| Posted: 07 May 2007 01:47:06 pm Post subject: |
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according to my teacher I was wrong.. the answer was 1/4.. (sorry for suggesting you were wrong)
but because she hadn't "taught" that yet she gave full credit just for the attempt..
So, after class I went and complained about all the crap on the hw..
she put: f(x)=2x2+7x; find f1(25)
so I put 2(25)2+7(25)
and she counted it wrong because I was supposed to be finding inverses, apparently she forgot the negative sign.. (but she still took off for it ...)
(there were tons of errors like that, she also worded "questions" poorly so you had to guess at what she meant.. and she is counting this as a test grade.. *shake fist*)
So how did you get 1/4? Please show all steps so I can learn
Last edited by Guest on 07 May 2007 01:47:36 pm; edited 1 time in total |
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simplethinker
snjwffl
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Joined: 25 Jul 2006
Posts: 700
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| Posted: 07 May 2007 02:20:45 pm Post subject: |
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4x²+4y²-24x+8y+39=0
4(x²-6x)+4(y²+2y)+39=0
4(x²-6x+9)-(4*9)+4(y²+2y+1)-(4*1)+39=0
4(x-3)²+4(y+1)²+39-36-4=0
4(x-3)²+4(y+1)²=1
(x-3)²+(y+1)²=1/4
here's a lot easier way to do cos(2x)²-cos(x)²=sin(x)²-sin(2x)²
the equation can be rearranged to:
cos(2x)²+sin(2x)²=sin(x)²+cos(x)²
and because of the identity sin(u)²+cos(u)²=1
cos(2x)²+sin(2x)²=1, and cos(x)²+sin(x)²=1,
so after plugging them in, you're left with 1=1
Last edited by Guest on 07 May 2007 02:24:23 pm; edited 1 time in total |
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garrett01001
Member
Joined: 21 Apr 2007
Posts: 135
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| Posted: 07 May 2007 02:33:14 pm Post subject: |
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thanks
factoring isn't my forte.. (obviously) |
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frenchcalc1
جان ألعريم
Active Member
Joined: 14 Mar 2007
Posts: 648
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| Posted: 07 May 2007 07:12:05 pm Post subject: |
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garrett01001 wrote: (there were tons of errors like that, she also worded "questions" poorly so you had to guess at what she meant.. and she is counting this as a test grade.. *shake fist*)
[post="103202"]<{POST_SNAPBACK}>[/post]
That reminds me of my physics teacher! He is a first generation Phillipino, and some of his sentences are so wrongly worded that at times I think he is stating something rather than asking it  He also takes off full credit for correcting his sentences on the test sheet :hmpf: |
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thornahawk
μολών λαβέ
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Joined: 27 Mar 2005
Posts: 569
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| Posted: 09 May 2007 06:27:04 am Post subject: |
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Well frenchcalc1, I don't think your teacher being first-gen Filipino has anything to do with his atrocious grammar and syntax. ;)
But that is already off-topic. XD
Any other questions, garrett01001? :)
thornahawk |
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garrett01001
Member
Joined: 21 Apr 2007
Posts: 135
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| Posted: 09 May 2007 02:22:25 pm Post subject: |
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none ATM..
I hate to boast, but I'm pretty good with math.. just not factoring.. and especially not good with stuff I haven't learned or even heard of.. |
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