How do you identify all vertical asymptotes for #f(x)=1-3/(x-3)#?

1 Answer
Jan 1, 2017

Answer:

vertical asymptote at x = 3

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: #x-3=0rArrx=3" is the asymptote"#

If you prefer you could consider f(x) as

#f(x)=(x-3)/(x-3)-3/(x-3)=(x-6)/(x-3)#

The end result is the same.
graph{(x-6)/(x-3) [-10, 10, -5, 5]}