# How do you find the domain and range of y=x^2+3x+1?

##### 1 Answer
May 2, 2017

$\text{D} : \left\{x \in \mathbb{R}\right\}$
$\text{R} : \left\{y \in \mathbb{R} | y \ge - 1.25\right\}$.

#### Explanation:

The domain and range are a set of all the possible values that a function can have.

Domain refers to the $x$-coordinate, and range refers to the $y$-coordinate.

For a parabola, no matter the values, the domain will always be $\text{D} : \left\{x \in \mathbb{R}\right\}$ (unless context is given).

As for range, the range is dependent on the $c$-value of the equation (only in vertex form).

This is because, if the $c$-value (again, only in vertex form) is $1$, then the parabola is moved up $1$ units. Meaning any value below $1$ is inadmissible.

Unfortunately, in this case (standard form), the $c$-value refers to the $y$-intercept. We can convert the equation to vertex form, or we can graph it and examine the parabola. We'll do that.

graph{x^2 + 3x + 1 [-3.895, 3.9, -1.948, 1.947]}

As you can see, the domain can be any $x$-value, while the range can only be values equal to or above the vertex's $y$-coordinate, $- 1.25$.

Therefore, the range is $\text{R} : \left\{y \in \mathbb{R} | y \ge - 1.25\right\}$.

Hope this helps :)