Question

I have not done asymtotes since January and I threw away my notes when school ended :/

x-3/x+2 how do you figure out the vertical and horizontal asymptote

Answer 1

Vertical:
Look for values of x that'll make the denominator equal zero. The denominator is x+2... find x that'll make it zero, that's your vertical asymptote, x = ...

Horizontal:
Look at the highest powers of x in the numerator and denominator... they are x and x. What is x/x equal to? That's your horizontal asymptote, y = ...

Answer 2

I remember a little of the horizontal, do I just see which one higher than the other?

Answer 3

Yes, basically. y = x/x = ...?

Answer 4

so x/x is equal to 1

Answer 5

Correct. H.A. is y=1.

Answer 6

when will the horizontal = 0 or none

Answer 7

H.A. will be y = 0 when the highest power of x on the top is smaller than the bottom. Eg \[\Large y = \frac{ x }{ x^2 }\]

H.A. won;'t exist when it's the other way around... highest power on the top is bigger than the bottom eg. \[\Large y = \frac{ x^2 }{ x }\] Well it will have a "slant" asymptote, but you may not have to worry about those yet.

Answer 8

I have this problem where its (x+6)(x-6)/x^2+9 would the H.A. 1 or none

Answer 9

well the top part has an x^2 term after you expand out (x+6)(x-6)

to find the H.A.
\[\Large y = \frac{ x^2 }{x^2 } = ...?\]