anonymous
  • anonymous
What are the vertical asymptotes of R(x) = 3x - 3 ______ ? x^2 - 4 A) x = 0 B) x = 1, and x = -1 C) x = 2, and x = -2 D) x = 3, and x = -3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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campbell_st
  • campbell_st
what values of x make the denominator zero, they are vertical asymptotes so you need to factor the quadratic denominator and then solve for x.
campbell_st
  • campbell_st
You should recognise the denominator is the difference of 2 squares.
anonymous
  • anonymous
@Michele_Laino

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Michele_Laino
  • Michele_Laino
the vertical asymptotes are located at points \(x\), such that the denominator is zero, namely: \(x^2-4=0\) so we have to solve this equation: \[\huge {x^2} - 4 = 0\]
Michele_Laino
  • Michele_Laino
hint: I can rewrite such equation as follows: \[\huge \left( {x - 2} \right)\left( {x + 2} \right) = 0\] now, please apply the product cancellation law, in order to write the solutions of the above equation
anonymous
  • anonymous
@Anaise
anonymous
  • anonymous
stil confused @Michele_Laino
Anaise
  • Anaise
long time no see, rip van pelt ;p
anonymous
  • anonymous
huh @Anaise can u help me
Michele_Laino
  • Michele_Laino
please solve this equations: \(x+2=0\), what is \(x\) ?
Michele_Laino
  • Michele_Laino
equation*
anonymous
  • anonymous
-2
Michele_Laino
  • Michele_Laino
that's right! Our first asymptote has this equation: \(x=-2\) Now, please solve this equation: \(x-2=0\), what is \(x\) ?
anonymous
  • anonymous
so the answer is C
Michele_Laino
  • Michele_Laino
that's right!

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