# What is the domain and range of f(x) =( x^2 - x - 6) / (x^2 + x - 12)?

Feb 19, 2016

Domain is all values except $x = - 4$ and $x = 3$ range is from $\frac{1}{2}$ to $1$.

#### Explanation:

In a rational algebraic function $y = f \left(x\right)$, domain means all values that $x$ can take. It is observed that in the given function $f \left(y\right) = \frac{{x}^{2} - x - 6}{{x}^{2} + x - 12}$, $x$ cannot take values where ${x}^{2} + x - 12 = 0$

Factorizing this becomes $\left(x + 4\right) \left(x - 3\right) = 0$. Hence domain is all values except $x = - 4$ and $x = 3$.

Range is values that $y$ can take. Although, one may have to draw a graph for this, but here as ${x}^{2} - x - 6 = \left(x - 3\right) \left(x + 2\right)$ and hence

$f \left(y\right) = \frac{{x}^{2} - x - 6}{{x}^{2} + x - 12} = \frac{\left(x - 3\right) \left(x + 2\right)}{\left(x + 4\right) \left(x - 3\right)} = \frac{x + 2}{x + 4}$

= $1 - \frac{2}{x + 4}$

and hence range is from $\frac{1}{2}$ to $1$.