How do you find the domain of $k (a) = \frac{a^{2}}{2 a^{2} + 3 a - 5}$ $k(a)=a^2/(2a^2+3a-5)$ ?

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Answer 1 · 2017-08-25T03:36:11

Domain of $f (a) = (- \infty, - \frac{5}{2}) \cup (- \frac{5}{2}, 1) \cup (1, + \infty)$ $f(a) = (-oo, -5/2)uu(-5/2, 1)uu(1, +oo)$

$k (a) = \frac{a^{2}}{2 a^{2} + 3 a - 5}$ $k(a) =a^2/(2a^2+3a-5)$

$= \frac{a^{2}}{(2 a + 5) (a - 1)}$ $= a^2/((2a+5)(a-1))$

Note that $k (a)$ $k(a)$ 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 $y$ .

How do you find the domain of k(a)=a^2/(2a^2+3a-5)k(a)=a22a2+3a−5k(a)=a^2/(2a^2+3a-5)?