How do you find the domain, range, and asymptotes for #4+5/(x+1)^2#?

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Can you give me a step-by-step on how you got your answer, so I can understand?

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Answer 1 · 2018-02-05T12:11:15

See below.

For the domain:

Notice if #x=-1# , #(x+1)^2=0#.

This is undefined ( division by zero ).

The function is therefore valid for all real #x#, #x!=-1#

We can write this in set notation as:

#{ x in RR : x!=-1}#

Or in interval notation as:

#(-00,-1)uu(-1,oo)#

Vertical asymptotes occur where the function is undefined. We have already found this to be at #bb(x=-1)#.

So the line #color(blue)(x=-1)# is a vertical asymptote:

as #x->+-oo# '#color(white)(88)5/(x+1)^2->0=>4+5/(x+1)^2#

#=4+0=4#

This means the function approaches #bb(4)# as x increases in both directions.

Therefore the line #color(blue)(y=4)# is a horizontal asymptote.

This is also the minimum value of the function.

as, #x->-1# , #color(white)(88)5/(x+1)^2->oo#

#:.#

#4+oo=oo#

Notice from the graph that this is vertical asymptote we found earlier.

So the range is:

#{y in RR : 4 < y < oo }#

#(4 , oo)#

GRAPH:

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