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Pre Calc. Can anyone help me?

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What's the question?
"Can anyone help me?"

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

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