How do you graph #f(x)=(2x^2+10x+12)/(x^2+3x+2)# using holes, vertical and horizontal asymptotes, x and y intercepts?

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Answer 1 · 2017-09-10T05:28:03

Vertical asymptote at #x=-1#, horizontal asymptote at #y=2#, hole at #x=-2#, #y#-intercept at #(0,6)# and #x#-intercept at #x=-3#.

When #x=0# #f(x)=(2x^2+10x+12)/(x^2+3x+2)=6#, so we have a #y#-intercept at #(0,6)#.

As #f(x)=(2x^2+10x+12)/(x^2+3x+2)#

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

= #(2(x+3)(x+2))/((x+1)(x+2))#

= #2(x+3)/(x+1)#

= #2(1+2/(x+1))#

= #2+4/(x+1)#

And when #y=0#, #2+4/(x+1)=0# or #4/(x+1)=-2# leading to #x=-3# and we have a #x#-intercept at #(-3,0)#.

Observe that although #x+2# cancels out #(2x^2+10x+12)/(x^2+3x+2)# is not defined for #x+2=0# i.e. #x=-2#. Hence we have a hole at #x=-2#.

Also for #x->-1#, from left #f(x)->-oo# and from right #f(x)->oo#, so we have a vertical asymptote at #x=-1#

Further, as #x->oo# or #x->-oo#, #f(x)->2# and hence we have a horizontal asymptote at #y=2#.

The graph using these points and trends appears as follows:

graph{(2x^2+10x+12)/(x^2+3x+2) [-9.79, 10.21, -4.24, 5.76]}