How do you graph #y=(x^2-4)/(x^2+3)# using asymptotes, intercepts, end behavior?

1 Answer
Narad T.
Oct 26, 2016

Only a horizontal asymptote #y=1#
Intercepts are #0,-4/3# , (2,0) and (-2,0)
minimum at #(0,-4/3)#
And limit #x->+-oo# is #y->1#

Explanation:

We start by finding the Domain
Here it is #RR# as the denominator is always positive
#x^2+3>0#
So there is no horizontal asymptote
The intercepts are #(0,-4/3)# on the y axis
and #(2,0) and (-2,0)# on the x axis
The we look for symmetry, #f(-x)=f(x)#
So thre is symmetry about the y axis
limit #y=1#
#x->-oo#

limit #y=1#
#x->oo#

So #y=1# is a horizontal asymptote
We can calculate the first derivative #y'=(2x(x^2+3)-2x(x^2-4))#
#=(2x^3+6x-2x^3+8x)/(x^2+3)=(14x)/(x^2+3)^2#
So there is a minimum at #(0,-4/3)#

graph{(x^2-4)/(x^2+3) [-8.89, 8.89, -4.444, 4.445]}

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