SOLUTION: asymptotes f(x)=(3x^2-7x-6)/(2x^2-18)

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Question 435914: asymptotes
f(x)=(3x^2-7x-6)/(2x^2-18)
Found 2 solutions by richard1234, stanbon:
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
The asymptotes will occur when the denominator is 0 and the numerator is nonzero.

The denominator is 0 when 2x%5E2+-+18+=+0 --> x+=+3, x+=+-3. However, when x = 3 we get f%283%29+=+0%2F0 which is indeterminate, and no asymptote occurs (there is a hole in the graph instead). Since f%28-3%29+%3C%3E+0, this is the only asymptote.

graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C%283x%5E2-7x-6%29%2F%282x%5E2-18%29%29

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
asymptotes
f(x)=(3x^2-7x-6)/(2x^2-18)
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Factor:
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f(x) = [3x^2-9x+2x-6]/[2(x-3)(x+3)]
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f(x) = [3x(x-3)+2(x-3)]/[2(x-3)(x+3)]
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f(x) = [(x-3)(3x+2)]/ [2(x-3)(x+3]
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Vertical asymptote at x = -3
Hole at x = 3
Horizontal asymptote: y = 3/2
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Cheers,
Cheers,
Stan H.