Question #004c3

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Answer 1 · 2017-11-09T18:29:32

vertical; at #x=2# and #x=3#

#x^2+7x+12 = (x+3)(x+4)#

#x^2-5x+6 = (x-3)(x-2)#

#f(x) = ((x+3)(x+4))/((x-3)(x-2))#

#n/0 =# undefined

therefore, asymptotes are at #x=2# and #x=3#

desmos.com/calculator

(the orange and green lines are #x=2# and #x=3#)

Answer 2 · 2017-11-09T18:37:44

See below.

The function will have a vertical asymptotes where it is undefined. So for the denominator equalling zero, we have:

#x^2-5x+6=0#

Factoring:

#(x-3)(x-2)=0=>x=3 or x=2#

The lines #x=3 and x=2# are vertical asymptotes.

For end behaviour, we have:

#x^2/x^2=1#

so

#lim_(x->oo)(x^2+7x+12)/(x^2-5x+6)=1#

#lim_(x->-oo)(x^2+7x+12)/(x^2-5x+6)=1#

So the line #y=1# is a horizontal asymptote.

Graph:

enter image source here