How do you graph #f(x)=(3x^2+2)/(x+1)# using holes, vertical and horizontal asymptotes, x and y intercepts?
1 Answer
V.A
H.A none
S.A
no
no holes
Explanation:

V.A are the zeros (undefined points) of the denominator
#x+1=0#
#x= 1#
vertical asymptotes are#x= 1# 
H.A if the degrees of the numerator is equal to the degree of denominator,
if the numerators degree > 1 + degree of denominator, there is a slant asymptote
degree of numerator is 2, degree of denominator is 1
#y=mx+b#
#(3x^2+2)/(x+1)#
#=3x+(3x+2)/(x+1)#
#=3x3+(5)/(x+1)#
#y=3x3#
slant asymptote is#y=3x3# 
#x# intercept is a point on the graph where#y=0#
#(3x^2+2)/(x+1)=0# no solution for#x inRR#
no#x# intercept 
#y# intercept is the point on the graph where#x=0#
#y=(3 * 0^2+2)/(0+1)#
#y=2#
#y# intercept is#(0,2)# 
no holes because the denominator doesn't cancel out