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

Question

Eli Assouline · Accepted Answer

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

#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)#

plot{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#.

Eva Abramson · Answer

undefined To find the domain of the function $ k(a) = \frac{a^2}{2a^2 + 3a - 5} $, you need to identify any values of $ a $ that would make the denominator equal to zero, as division by zero is undefined.

1. First, factor the denominator $ 2a^2 + 3a - 5 $.
2. Then, set each factor equal to zero and solve for $ a $.
3. The domain of the function will be all real numbers except for the values of $ a $ that make the denominator zero.

Once you find the values of $ a $ that make the denominator zero, you exclude them from the domain, and the remaining values of $ a $ will be the domain of the function.