How do you find the domain of #k(a)=a^2/(2a^2+3a-5)#?

1 Answer
Aug 25, 2017

Answer:

Domain of #f(a) = (-oo, -5/2)uu(-5/2, 1)uu(1, +oo)#

Explanation:

#k(a) =a^2/(2a^2+3a-5)#

#= a^2/((2a+5)(a-1))#

Note that #k(a)# is defined for all #a in RR# except where #(2a+5)(a-1) =0#

I.e. where #a=-5/2 or 1#

Hence #k(a)# is defined #forall a in RR: a!= {-5/2, 1}#

#:.# the domain of #k(a)# is all #a in RR: a!= {-5/2, 1}#

Or in interval notation: #(-oo, -5/2)uu(-5/2, 1)uu(1, +oo)#

graph{x^2/((2x+5)(x-1)) [-10, 10, -5, 5]}

This is demonstrated by the graph of #k(a)# below - Where the axes are #a# and #k(a)# replacing the conventional #x# and #y#.