Question

What are the vertical asymptotes, horizontal asymptotes, slant asymptotes, and the holes of this equation:

(x^2-9x+20)/(2x^2-8x)

Answer 1

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

Answer 2

a hole is where the bottom gets canceled out by a like facotr on the top, so factor your top as well

Answer 3

(x-5)(x-4)
---------
2x(x-4)

Answer 4

horizontal asymp are the most fun

Answer 5

there is a hole at x=4; and a VA at x = 0

Answer 6

are you sure that the horizontal asymptote is 1/2 because i thought it was 2

Answer 7

its 1/2

Answer 8

im sure; if its what you wrote up top then:

x^2-9x+20 1+.....
----------- = -------- = 1/2
2x^2-8x 2+.....

Answer 9

you take the limit as x goes to infinity

Answer 10

divide top and bottom by x^2 to clear things up and yeah, you end up with 1/2

Answer 11

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

Answer 12

divide everything by the highest degree variable in the bottom.... anything left stuck with an x in its bottom gets thrown in the trash

Answer 13

ok and what is the slant?

Answer 14

there is no slantl the slant is what happens the degree of the top is heavier than the degree of the bottom

Answer 15

(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

Answer 16

the slant is when you do long division. so what is the slant?

Answer 17

its a line as that the curve gets closer to as x gets closer to infinity

Answer 18

there.....is.....NO .....slant.......to .....your.....problem.....