Question

Find the horizontal asymptote of y= 1/x-3 + 2.
I just don't get the whole asymptote thing.

Answer 1

You have written: \(y = \dfrac{1}{x} - 3 + 2\)

I'm guessing this is not your intent.

Give it another go and remember your Order of Operations.

Answer 2

Sorry I meant \[y= \frac{ 1 }{ x-3 } + 2\]

Answer 3

Consider the degree of the numerator and denominator separately. What say you?

Answer 4

Num deg = 0, denom. deg = 1

Answer 5

So, since the numerator is a lower degree than the denominator, then there is a horizontal asymptote y= 0.

Answer 6

So do I plug that in?

Answer 7

\[0 =\frac{ 1 }{x-3}+ 2\]

Answer 8

????????

Answer 9

You see a horizontal asymptote. Why are you substituting anything?

\(y = \dfrac{1}{x-3}\) has a horizontal asymptote at y = 0

Move it up with a translation of 2.

\(y = \dfrac{1}{x-3} + 2\) has a horizontal asymptote at y = 2

Answer 10

OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOH!

Answer 11

THANK YOU!

Answer 12

You can alsow rewrite, by adding the fractions.

\(y = \dfrac{1}{x-3} + 2 = \dfrac{2x-5}{x-3}\). The asymptote at y = 2 is more obvious in this form.