# How do you find the domain and range of f(x) =x^2+8x+15?

Feb 20, 2018

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

Range: $\left[- 1 , \infty\right)$

#### Explanation:

Given:

$f \left(x\right) = {x}^{2} + 8 x + 15$

First note that in common with any polynomial, the (implicit) domain of $f \left(x\right)$ is the whole of the real numbers $\left(- \infty , \infty\right)$.

We can find the range by completing the square:

${x}^{2} + 8 x + 15 = {x}^{2} + 8 x + 16 - 1 = {\left(x + 4\right)}^{2} - 1$

Note that ${\left(x + 4\right)}^{2} \ge 0$ for any real value of $x$, taking the value $0$ when $x = - 4$. So $f \left(x\right)$ takes its minimum value $- 1$ when $x = - 4$.

Then since $f \left(x\right)$ is continuous and increases without limit as $x \to \pm \infty$.

So the range is $\left[- 1 , \infty\right)$.

Another way of finding the range is to set $y = f \left(x\right)$ then solve for $x$...

Given:

$y = {x}^{2} + 8 x + 15 = {\left(x + 4\right)}^{2} - 1$

Add $1$ to both ends to get:

$y + 1 = {\left(x + 4\right)}^{2}$

Take the square root of both sides, allowing for both possibilities to get:

$x + 4 = \pm \sqrt{y + 1}$

Subtract $4$ from both sides to get:

$x = - 4 \pm \sqrt{y + 1}$

So for any $y \ge - 1$, there is at least one value of $x$ such that $f \left(x\right) = y$. In other words, any $y \in \left[- 1 , \infty\right)$ is part of the range, and no other values are.