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)#
#=3x-3+(5)/(x+1)#
#y=3x-3#
slant asymptote is#y=3x-3# -
#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
