# What is the domain and range of y=(x^2-x-1) / (x+3)?

Feb 23, 2018

Domain: $\left(- \infty , - 3\right) \cup \left(- 3 , \infty\right)$

Range: $\left(- \infty , - 2 \sqrt{11} - 7\right] \cup \left[2 \sqrt{11} - 7 , \infty\right)$

#### Explanation:

The domain is all values of $y$ where $y$ is a defined function.

If the denominator is equal to $0$, the function is typically undefined. So here, when:

$x + 3 = 0$, the function is undefined.

Therefore, at $x = - 3$, the function is undefined.

So, the domain is stated as $\left(- \infty , - 3\right) \cup \left(- 3 , \infty\right)$.

The range is all possible values of $y$. It is also found when the discriminant of the function is less than $0$.

To find the discriminant ($\Delta$), we must make the equation a quadratic equation.

$y = \frac{{x}^{2} - x - 1}{x + 3}$

$y \left(x + 3\right) = {x}^{2} - x - 1$

$x y + 3 y = {x}^{2} - x - 1$

${x}^{2} - x - x y - 1 - 3 y = 0$

${x}^{2} + \left(- 1 - y\right) x + \left(- 1 - 3 y\right) = 0$

This is a quadratic equation where $a = 1 , b = - 1 - y , c = - 1 - 3 y$

Since $\Delta = {b}^{2} - 4 a c$, we can input:

$\Delta = {\left(- 1 - y\right)}^{2} - 4 \left(1\right) \left(- 1 - 3 y\right)$

$\Delta = 1 + 2 y + {y}^{2} + 4 + 12 y$

$\Delta = {y}^{2} + 14 y + 5$

Another quadratic expression, but here, since $\Delta \ge 0$, it is an inequality of the form:

${y}^{2} + 14 y + 5 \ge 0$

We solve for $y$. The two values of $y$ we get will be the upper and lower bounds of the range.

Since we can factor $a {y}^{2} + b y + c$ as $\left(y - \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a}\right) \left(y - \frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a}\right)$, we can say, here:

$a = 1 , b = 14 , c = 5$. Inputting:

$\frac{- 14 \pm \sqrt{{14}^{2} - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

$\frac{- 14 \pm \sqrt{196 - 20}}{2}$

$\frac{- 14 \pm \sqrt{176}}{2}$

$\frac{- 14 \pm 4 \sqrt{11}}{2}$

$\pm 2 \sqrt{11} - 7$

So the factors are $\left(y - \left(2 \sqrt{11} - 7\right)\right) \left(y - \left(- 2 \sqrt{11} - 7\right)\right) \ge 0$

So $y \ge 2 \sqrt{11} - 7$ and $y \le - 2 \sqrt{11} - 7$.

In interval notation we can write the range as:

$\left(- \infty , - 2 \sqrt{11} - 7\right] \cup \left[2 \sqrt{11} - 7 , \infty\right)$