anonymous
  • anonymous
I need help please.... Find the domain of the rational function F(x)=-2(x^2-4)/3(x^2+4x+4) I have factored them correctly but it said after I factored them I shouldn't reduce any common factors and should find the domain of the function using its factored form and that I need to state the domain in interval notation. I am lost and do not know how to work this out apparently
Mathematics
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SOLVED
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chestercat
  • chestercat
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nenadmatematika
  • nenadmatematika
\[x^2+4x+4\neq0 \] \[(x+2)^2\neq0\] so the domain interval is:
nenadmatematika
  • nenadmatematika
\[x^2+4x+4\neq0 \] \[(x+2)^2\neq0\] so the domain interval is: \[x \in(-\infty,-2)\cup(-2,+\infty)\]
anonymous
  • anonymous
R-{-2}

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anonymous
  • anonymous
Check your nos nena..
nenadmatematika
  • nenadmatematika
to check what?
anonymous
  • anonymous
I am not understanding...i was told to find the domain of the function using the factored form F(x)=[-2(x-2)(x+2)]/[3(x+2)(x+2)]....how do I work this out
anonymous
  • anonymous
Sorry my bad
anonymous
  • anonymous
Once factored you must use fact the Denominator cannot be zero for any real no
nenadmatematika
  • nenadmatematika
it's ok...:D kcbrosell what part you don't understand?
anonymous
  • anonymous
all of it I am failing this class big time...I dont understand all the factoring or how to put it into and equation and so on
nenadmatematika
  • nenadmatematika
every time you want to find the domain of the function which is given in fraction form denominator must not be equal to zero, so when you factor x^2+4x+4 you get the result I wrote.....
anonymous
  • anonymous
I am not understanding how you get to the result u wrote
nenadmatematika
  • nenadmatematika
well it's a basic binomial formula which you must know if you want to do some serious math in the future....
anonymous
  • anonymous
I got to this point x^2-4=(x-2)(x+2) & x^2+4x+4=(x^2+2)
nenadmatematika
  • nenadmatematika
you got wrong x^2+2x+4...it is (x+2)^2
anonymous
  • anonymous
o crud...so after I get that worked out do I have to put them into another equation
anonymous
  • anonymous
|dw:1328644954475:dw|
anonymous
  • anonymous
if terms like (x+2) cancel, there would normally be a "hole" in the graph...but note that there is still a (x+2) in the denominator, so you must set it equal to zero and solve....this is the value that won't work....so the domain is all of the other numbers....the numbers less than -2 and the numbers greater than -2.
anonymous
  • anonymous
so then my answer would be (-oo, -2) (-2, oo)
anonymous
  • anonymous
yes.
anonymous
  • anonymous
so it all has to do with your neg and pos and how they factor out
anonymous
  • anonymous
think about what would make the denominator equal to zero...you are correct you often times must factor first to be able to tell.
anonymous
  • anonymous
okay i thinki understand it a little bit better now thank you for your help
anonymous
  • anonymous
:)

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