Question

find vertical & horizontal asymptote of f(x)= x^2-3/(x-2)(x-1)

Answer 1

a vertical asymptote is where the function the value of f does not exist meaning denominator equals 0. So what do you suppose it is?

Answer 2

basically set (x-2)(x-1) = 0

Answer 3

what do i do with the top x^2 -3?

Answer 4

well it is used but only the x^2 portion. a horizontal asymptote is found using the degrees of the numerator and denominator if the degree of the numerator is greater then there is no horizontal asymptote, if the denominator is greater then the asymptote is x = 0, if it is a tie then it is the ratio of the coefficients

Answer 5

i will draw this out real quick with an example

Answer 6

thank you :)

Answer 7

make sense ... thank you

Answer 8

so you are basically looking at the powers of the first terms in both top and bottom

Answer 9

so if the denominator is greater than there is no H.A

Answer 10

the top

Answer 11

correct but in this case that is not true. It is a tie

Answer 12

factor the bottom first and you will see you got x^2/x^2

Answer 13

ohh i see lol

Answer 14

yes so your H.A. is at y = 1 and verticals should be at x - 2 and x = 1