How do you find the asymptotes for # f(x) = (2x-2) /( (x-1)(x^2 + x -1))#?
1 Answer
Explanation:
#"the first step is to factorise and simplify f(x)"#
#f(x)=(2cancel((x-1)))/(cancel((x-1))(x^2+x-1))=2/(x^2+x-1)# 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+x-1=0" using the quadratic formula"#
#x=(-1+-sqrt(1+4))/2#
#rArrx=-1/2+-1/2sqrt5#
#rArrx~~-1.62" and " x~~0.62" are the asymptotes"# Horizontal asymptotes occur as
#lim_(xto+-oo),f(x)toc" ( a constant)"# divide terms on numerator/denominator by the highest power of x, that is
#x^2#
#f(x)=(2/x^2)/(x^2/x^2+x/x^2-1/x^2)=(2/x^2)/(1+1/x-1/x^2)# as
#xto+-oo,f(x)to0/(1+0-0)#
#rArry=0" is the asymptote"#
graph{2/(x^2+x-1) [-10, 10, -5, 5]}
