SOLUTION: We need to find the verticle and horizontal asymptote of this: y=((3x-6)/(2x^2 + 6x - 80))

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Question 295350: We need to find the verticle and horizontal asymptote of this:
y=((3x-6)/(2x^2 + 6x - 80))
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
We need to find the verticle and horizontal asymptote of this:
y=((3x-6)/(2x^2 + 6x - 80))
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Horizontal?
Procedure:
Find the term with the highest power of the variable.
That is 2x^2 in the denominator
and 0x^2 in the numerator.
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Then y = 0x^2/2x^2 = 0/2 = 0 is the horizontal asymptote.
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Vertical asymptote?
Procedure:
Factor the denominator:
2(x^2+3x - 40)
2(x^2+8x-5x-40)
2(x(x+8)-5(x+8))
= 2(x+8)(x-5)
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Solve 2(x+8)(x-5) = 0
x = -8 or x = 5
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You have vertical asymptotes at x = -8 and at x = 5
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Cheers,
Stan H.
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