# How do you find the domain and range of sqrt(x^2-5x-14)?

Jul 20, 2017

$\left(- \infty , - 2\right] U \left[7 , \infty\right)$

#### Explanation:

Domain is the possible values for $x$.

The domain for the parent function , $\sqrt{x}$ is from $0$ to $\infty$, or $\left[0 , \infty\right)$
graph{y=sqrtx}

Notice how $0$ is included. That's because $\sqrt{0}$ is the lowest number that is possible. Any number smaller than $0$ is negative, and a square root of a negative number craetes an unreal number ($i$), which we can't graph.

Long story short, we can't take the square rot of a negative number, so we stop at $0$

Let's factor the quadratic and see the roots for $x$:

${x}^{2} - 5 x - 14$

We need to find $2$ numbers that multiply to $- 14$ and add to $- 5$

$\textcolor{w h i t e}{\times} + - 5$
$\textcolor{w h i t e}{+} \times - 14$
.........................
$\pm 1 \times \pm 14$
$\textcolor{red}{\pm 2 \times \pm 7}$

$- 7 + 2$ gives us $- 5$, and $- 7 \times 2$ equals $- 14$

So, now we have $\sqrt{\left(x + 2\right) \left(x - 7\right)}$

So, the rule with square roots is that the factors must be equal to or larger than $0$

Let's solve for these roots and see what value of $x$ makes everything equal $0$

factor 1
$x + 2 = 0$

$x = - 2$

factor 2
$x - 7 = 0$

$x = 7$

Now we know that if $x$ equals $- 2$ or $7$, we're taking a square root of $0$. So, if $x$ equals anything smaller than $7$ or larger than $- 2$, we would be taking a square root of a negative number.

So, our domain is "anything smaller than $- 2$ and anything larger than $7$", or $\left(- \infty , - 2\right] U \left[7 , \infty\right)$

To check our work, let's graph the equation:
graph{y=sqrt((x+2)(x-7))}

Yep! We were right.

The graph has no issues until $- 2$ to $7$. From those points, the graph is not possible