Question

How do I find the asymptotes of -3/x and 11/x-9+9?

without graphing?

Answer 1

wait you have,

\(\large\color{black}{ y=\frac{\LARGE -3 }{\LARGE x} }\) and \(\large\color{black}{ y=\frac{\LARGE 11 }{\LARGE x-9} +9 }\) correct?

Answer 2

yes

Answer 3

the asymptote will be the value that x can't equal.
What value can't x equal, when you have: \(\large\color{black}{ y=\frac{\LARGE -3 }{\LARGE x} }\) ?

Answer 4

( number/0 is undefined)

Answer 5

0

Answer 6

yes.

Answer 7

so x and y asymptotes are 0?

Answer 8

in this case yes.

Answer 9

to compare -3/x and 1/x, would it just be that -3/x is increasing while 1/x is decreasing?

Answer 10

I am still typing, I dissapeared for some reason

Answer 11

i know, the site is messed up lately

Answer 12

\(\large\color{blue}{ y=\frac{\LARGE -3 }{\LARGE x} }\)

x=0 is the vertical asymptote, b/c:
it y will be completely different as x approaches 0 from left and right side (of the y axis)

when you plug in very small negative numbers you get a (-3/-0.001)
which becomes (3/0.001) and that is 3000.

NOW LETS,
approach zero (from left side), a bit closer,
(-3/-0.00001) >>> 3/0.00001 >>> 3000000

so as x goes closer and close to zero from the left (of the y axis), the y is infinity.

Answer 13

So, the opposite is for x approaching zero from the right side.
\(\large\color{blue}{ y=\frac{\LARGE -3 }{\LARGE x} }\)

(-3 / 0.01) >>> -300

(-3 / 0.0001) >>> -30,000

(-3 / 0.000001) >>> -3,000,000

Answer 14

Same thing, but the other direction, as x gets closer and closer to zero from the right of the x-axis, the y value becomes a very large negative i.e. \(\large\color{blue}{-\infty }\)