anonymous
  • anonymous
Pre Calc. Can anyone help me?
Mathematics
katieb
  • katieb
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3psilon
  • 3psilon
What's the question?
anonymous
  • anonymous
"Can anyone help me?"
anonymous
  • anonymous
YES! :D

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anonymous
  • anonymous
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anonymous
  • anonymous
\[f(x) = \frac{ 2 }{ x^2-2x-3 }=\frac{ 2 }{ (x+1)(x-3) }\]
anonymous
  • anonymous
okay
anonymous
  • anonymous
That means the x cannot be -1, 3 so the domain (-infi, -1)U(-1, 3)U(3, infi)
anonymous
  • anonymous
or x <-1, -13
anonymous
  • anonymous
okay
anonymous
  • anonymous
for x-intercept: set f(x)=0, then: \[0=\frac{ 2 }{ (x+1)(x-3) }\] since no x exist that satisfy the above equation, there is no x-intercept
anonymous
  • anonymous
for y-intercept: set x = 0, then: \[f(0) = \frac{ 2 }{ (0+1)(0-3) }=\frac{ -2 }{ 3 }\] so the y-intercept is at the point (0, -2/3)
anonymous
  • anonymous
To get the horizontal asymptote: divide every individual block in the equation by the highest power of x, which is x^2
anonymous
  • anonymous
no x inct? no range?
anonymous
  • anonymous
o lol, forgot about range and there's no x-int as I've explained
anonymous
  • anonymous
okay
anonymous
  • anonymous
sorry, I need to a lecture now, but here is the graph for reference. good luck :D
anonymous
  • anonymous
okay.
anonymous
  • anonymous
anyone?

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