How do you graph #f(x)=(-x-1)/(x+2)# using holes, vertical and horizontal asymptotes, x and y intercepts?
1 Answer
see explanation.
Explanation:
#color(blue)"Asymptotes"# 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+2=0rArrx=-2" is the asymptote"# Horizontal asymptotes occur as
#lim_(xto+-oo),f(x)toc" ( a constant)"# divide terms on numerator/denominator by x
#f(x)=(-x/x-1/x)/(x/x+2/x)=(-1-1/x)/(1+2/x)# as
#xto+-oo,f(x)to(-1-0)/(1+0)#
#rArry=-1" is the asymptote"# Holes occur if there is a duplicate factor on the numerator/denominator. This is not the case here hence there are no holes.
#color(blue)"Intercepts"#
#x=0toy=-1/2rArr(0,-1/2)larr"y-intercept"#
#y=0to-x-1=0tox=-1rArr(-1,0)#
graph{(-x-1)/(x+2) [-10, 10, -5, 5]}
