Question

Another question about asymptotes..

Answer 1

\[f(x)=\frac{ x-x^2 }{ 2-3x+x^2 }\]

Answer 2

ok so a vertical asymptote would be anything that makes the denominator zero

Answer 3

which would be 1 and 2

Answer 4

Yes

Answer 5

my book says only 2 is a VA

Answer 6

For 1, the equation becomes 0/0 which is not determined

Answer 7

ok so thats illegal and i cant do 1

Answer 8

yep...

Answer 9

ok this is some serious stuff lol, so many little things you have to watch out for :D

Answer 10

it's because if we try to evaluate the limit x-> 1
\[\lim_{x\to 1}\frac{x-x^2}{2-3x+x^2}\]
\[\lim_{x\to 1}\frac{x(1-x)}{(x-1)(x-2)}\]
\[\lim_{x\to 1}\frac{x\cancel{(1-x)}}{\cancel{(x-1})(x-2)}=\frac{-x}{x-2}=1\]
limit exists at x=1, that's why it's not a vertical asymptote

Answer 11

is @AbhimanyuPudi a good way to look at it?

Answer 12

plug it into origional function and check? or will that mess me up?

Answer 13

as x takes values very very near to 1, f(x) takes values very near to 1, as shown by ash
it doesn't go to infinity. hence only x=2 is VA

Answer 14

hmm, ok so i need to run limit checks on vertical asymptotes too?

Answer 15

yes.

Answer 16

ugh

Answer 17

or if you have confusion, why dont you just simplify in the beginning,,
like @ash2326 did,,

you can see 1 is not a part of the domain,, so you may thus cancel out x-1 from both num and denom..
now you can only have x=2..