How do you find the vertical, horizontal or slant asymptotes for #y = (x-3)/(x-2)#?
Vertical Asymptote: x= 2
Horizontal Asymptote: y= 1
No slant asymptote.
y= (x-3) / (x-2)
Non-permissible Value: x=2
If x=2, the denominator would be zero, and it can never be zero because any number divided by 0 will be undefined.
When x=2 and you plug it back into the original equation, the top has a value while the bottom is zero:
(2-3)/ (2-2)= (-1)/ (0)
That means there is a Vertical Asymptote.
The Horizontal Asymptote is y=1 because the top degree is equal to the bottom degree and their leading coefficients are both 1.
(1x) / (1x) = 1
Therefore the HA: y=1; now since a HA exists, a slant asymptote cannot at the same time.