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

Aug 25, 2017

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

#### Explanation:

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

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

Note that $k \left(a\right)$ is defined for all $a \in \mathbb{R}$ except where $\left(2 a + 5\right) \left(a - 1\right) = 0$

I.e. where $a = - \frac{5}{2} \mathmr{and} 1$

Hence $k \left(a\right)$ is defined $\forall a \in \mathbb{R} : a \ne \left\{- \frac{5}{2} , 1\right\}$

$\therefore$ the domain of $k \left(a\right)$ is all $a \in \mathbb{R} : a \ne \left\{- \frac{5}{2} , 1\right\}$

Or in interval notation: $\left(- \infty , - \frac{5}{2}\right) \cup \left(- \frac{5}{2} , 1\right) \cup \left(1 , + \infty\right)$

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

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