# What is the domain and range of y= (4x) / (x^2 + x - 12) ?

Jun 9, 2018

The domain is $x \in \left(- \infty , - 4\right) \cup \left(- 4 , 3\right) \cup \left(3 , + \infty\right)$. The range is $y \in \mathbb{R}$

#### Explanation:

The denominator must be $\ne 0$

Therefore,

${x}^{2} + x - 12 \ne 0$

$\left(x + 4\right) \left(x - 3\right) \ne 0$

$x \ne - 4$ and $x \ne 3$

The domain is $x \in \left(- \infty , - 4\right) \cup \left(- 4 , 3\right) \cup \left(3 , + \infty\right)$

To find the range, proceed as follows

$y = \frac{4 x}{{x}^{2} + x - 12}$

$\implies$, $y \left({x}^{2} + x - 12\right) = 4 x$

$\implies$, $y {x}^{2} + y x - 4 x - 12 y = 0$

In order for this equation to have solutions, the discriminant $\ge 0$

Therefore,

$\Delta = {\left(y - 4\right)}^{2} - 4 y \cdot \left(- 12 y\right)$

$= {y}^{2} + 16 - 8 y + 48 {y}^{2}$

$= 49 {y}^{2} - 8 y + 16$

$\forall y \in \mathbb{R} , \left(49 {y}^{2} - 8 y + 16\right) \ge 0$

as $\delta = {\left(- 8\right)}^{2} - 4 \cdot 49 \cdot 16 > 0$

The range is $y \in \mathbb{R}$

graph{(4x)/(x^2+x-12) [-25.66, 25.65, -12.83, 12.84]}