Question

What are the asymptotes and removable discontinuities, if any, of `f(x)=((3x -6)/(x^2 -5x + 6))`?

Answer 1

removable discontinuity at x = 2
vertical asymptote x = 3
horizontal asymptote y = 0

Explanation:

First step is to factor the function.

`( 3(x - 2 ))/((x - 3 )(x - 2 )) =( 3cancel((x-2)))/((x-3)cancel((x-2))`

There is a removable discontinuity at x = 2 .

f(x) is now f(x) `= 3/(x-3)`

Vertical asymptotes occur as the denominator tends to zero. To find the equation let the denominator equal zero.

solve : x - 3 = 0 → x = 3 is the asymptote

Horizontal asymptotes occur as `lim_(x→±∞) f(x) → 0`

divide all terms by x

`(3/x)/(x/x - 3/x) = (3/x)/(1- 3/x)`

As x → ∞ , `3/x → 0 , y = 0" is the asymptote "`

Here is the graph of the function.
graph{(3x-6)/(x^2-5x+6) [-10, 10, -5, 5]}