What are the vertical asymptotes, horizontal asymptotes, slant asymptotes, and the holes of this equation: (x^2-9x+20)/(2x^2-8x)

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What are the vertical asymptotes, horizontal asymptotes, slant asymptotes, and the holes of this equation: (x^2-9x+20)/(2x^2-8x)

Mathematics
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since the bottom and top degrees are the same; the HA is the qoutient of the top leading coefficient and the bottom leading coefficint HA = 1/2 Vas are where your bottom goes to zero; so factor it out and check for roots
a hole is where the bottom gets canceled out by a like facotr on the top, so factor your top as well
(x-5)(x-4) --------- 2x(x-4)

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Other answers:

horizontal asymp are the most fun
there is a hole at x=4; and a VA at x = 0
are you sure that the horizontal asymptote is 1/2 because i thought it was 2
its 1/2
im sure; if its what you wrote up top then: x^2-9x+20 1+..... ----------- = -------- = 1/2 2x^2-8x 2+.....
you take the limit as x goes to infinity
divide top and bottom by x^2 to clear things up and yeah, you end up with 1/2
since the degrees are the same you just take the coeiffient of the leading exponent term in the numerator over the coefficent of the leading exponent term in the denominator
divide everything by the highest degree variable in the bottom.... anything left stuck with an x in its bottom gets thrown in the trash
ok and what is the slant?
there is no slantl the slant is what happens the degree of the top is heavier than the degree of the bottom
(x^3+1)/(x^2+1) would have a slant since the degree of the top is one more than the degree of the bottm
the slant is when you do long division. so what is the slant?
its a line as that the curve gets closer to as x gets closer to infinity
there.....is.....NO .....slant.......to .....your.....problem.....

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