How do you graph #f(x)=(3x^2+2)/(x+1)# using holes, vertical and horizontal asymptotes, x and y intercepts?

1 Answer
Feb 18, 2018

V.A #x=-1#
H.A none
S.A #y=3x-3#
no #x#-intercept
#y#-intercept is #(0,2)#
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

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