Question

find vertical and horizontal asymptotes of a curve: y=(x^(2)+1)/(5x^(2)-44x-9)...please show steps thanks in advance: )

Answer 1

is it really
\[y=\frac{x^2+1}{5x^2-44x-9}\]

Answer 2

yes satellite73

Answer 3

vertical asymptote: set denominator = 0 and solve

Answer 4

by some miracle this actually factors as
\[(5x+1)(x-9)=0\] so vertical asymptotes are
\[x=-\frac{1}{5}\] and
\[x=9\]

Answer 5

ok so would be lik 5x^2-44x-9=0 and it would be like this?

Answer 6

horizontal asymptote much easier. the degrees are the same top and bottom (both 2) so take
\[y=\frac{\text{leading coefficient}}{\text{leading coefficient}}\]

Answer 7

ok so you factor it out right?

Answer 8

1/9?

Answer 9

yes that is right. put
\[5x^2-44x-9=0\] and solve for x

Answer 10

i mean 1/5

Answer 11

nope you factored wrong

Answer 12

still wrong. it is -1/5 or 9

Answer 13

wait for horizontal?

Answer 14

no no those are vertical. vertical where denominator is zero

Answer 15

the leading coeffeicent would be 1/5

Answer 16

the ratio of leading coefficients is 1/5 yes. so that is horizontal asymptote. you got it

Answer 17

yes but i meant to give answer for the horizontal section

Answer 18

yes it is
\[y=\frac{1}{5}\]

Answer 19

yes thanks soo much!! : )

Answer 20

yw