For what values of x, if any, does #f(x) = 1/((x-3)(x^2+4)) # have vertical asymptotes?
1 Answer
Aug 17, 2016
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 values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.
solve:
#x^2+4=0rArrx^2=-4" has no real solutions"#
and hence there are no vertical asymptotes.solve:
#x-3=0rArrx=3" is the asymptote"#
graph{(1/((x-3)(x^2+4))) [-10, 10, -5, 5]}
