How do you find the asymptotes for #y=2/(x-1) + 1#?
1 Answer
Jan 1, 2017
vertical asymptote at x = 1
horizontal asymptote at y = 1
Explanation:
The denominator of y cannot be zero as this would make y 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-1=0rArrx=1" is the asymptote"# You might prefer to consider y as.
The result is the same.
Horizontal asymptotes occur as
#lim_(xto+-oo),ytoc" ( a constant)"# divide terms on numerator/denominator by x
#y=(2/x)/(x/x-1/x)+1=(2/x)/(1-1/x)+1# as
#xto+-oo,yto0/(1-0)+1#
#rArry=1" is the asymptote"#
graph{(x+1)/(x-1) [-10, 10, -5, 5]}
