How do you graph f(x)=4/(x-1)+1 using holes, vertical and horizontal asymptotes, x and y intercepts?
1 Answer
see explanation.
Explanation:
We can express f(x) as a single rational function.
rArrf(x)=4/(x-1)+(x-1)/(x-1)=(x+3)/(x-1)
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-1=0rArrx=1" 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+3/x)/(x/x-1/x)=(1+3/x)/(1-1/x) as
xto+-oo,f(x)to(1+0)/(1-0)
rArry=1" is the asymptote" Holes occur when there is a duplicate factor on the numerator/denominator. This is not the case here, hence there are no holes.
color(blue)"Approaches to vertical asymptote"
lim_(xto1^-)=- oo" and " lim_(xto1^+)=+oo
color(blue)"Intercepts"
x=0tof(0)=3/(-1)=-3larr" y-intercept"
y=0tox+3=0tox=-3larr" x-intercept"
graph{(x+3)/(x-1) [-10, 10, -5, 5]}
