#g(x)=(x^2 + 4x - 5) / (x^2 + 7x + 10)=((x-1)(x+5))/((x+2)(x+5))#
Domain
#x!=-2 and x!=-5# as this would make the denominator #0#
As long as #x!=-5# we may cancel out the #(x+5)#'s
We get #g(x)=(x-1)/(x+2)#
There is a discontinuity at #x=-2 and x=-5#
At the #x=-5# discontinuity we may get as close as we want, because value #!=5# would make #(x+5)#'s cancel, and near this point the value of #g(x)# would be near #2#, or as we say:
#lim_(x->-5^+) g(x)=lim_(x->-5^-) g(x)=2# but #g(-5)# is undefined
Vertical asymptotes
#x=-2#, as we can say
#lim_(x->-2^+) g(x)=-oo# and #lim_(x->-2^-) g(x)=+oo#
Horizontal asymptote
As #x# gets larger and larger, the #-1# and #+2# in
#g(x)=(x-1)/(x+2)#matter less and less, so we can say:
#lim_(x->oo) g(x)=lim_(x->-oo) g(x)=1#
graph{(x^2+4x-5)/(x^2+7x+10) [-25.66, 25.66, -12.83, 12.81]}
