How do you find the domain for #R(x) = (x^2 + x - 12)/(x^2 - 4)#?

1 Answer
George C.
Jan 13, 2016

Find that #R(x)# is well defined for all #x# except #+-2#, so the domain of #R(x)# is #(-oo, -2) uu (-2, 2) uu (2, oo)#.

Explanation:

Both the numerator and denominator are well defined for any Real value of #x#. So the quotient #R(x)# will be well defined except when the denominator is zero.

#x^2-4 = 0# when #x = +-sqrt(4) = +-2#

So the domain of #R(x)# is #(-oo, -2) uu (-2, 2) uu (2, oo)#

Note that #x^2+x-12 = (x+4)(x-3)#, so the numerator is non-zero when #x = +-2#. As a result, #R(x)# has vertical asymptotes at #x=+-2# ...

graph{(x^2+x-12)/(x^2-4) [-20, 20, -10, 10]}

Related questions
See all questions in Domain and Range of a Function
Impact of this question
1552 views around the world
You can reuse this answer
Creative Commons License