# How do you find the domain and range and determine whether the relation is a function given :x=2y^2-3?

Sep 20, 2017

See below.

#### Explanation:

For y as a function of x.

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

Rearrange;

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

Divide by 2:

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

Taking roots of both sides:

$y = \sqrt{\frac{x + 3}{2}}$

For positive root:

For the domain we must make sure the numerator is $\ge 0$, since we are only interested in real numbers.

The numerator is $\ge 0$ for $x \ge - 3$

So the domain is $\left\{x \in \mathbb{R} | x \ge - 3\right\}$

The range for the positive root is:

As
$x \to + \infty$
$y \to + \infty$

So the range is:

$\left\{y \in \mathbb{R} | y \ge 0\right\}$

For negative root the domain is the same for real numbers. The range is in the reverse direction:

$\left\{y \in \mathbb{R} | y \le 0\right\}$

The determination of a function.

For positive root only a function is defined.
For negative root only a function is defined.

For $\pm \sqrt{\frac{3 + x}{2}}$

This is not a function. A function is only defined if an element in a domain is mapped onto one and one only element in a range. In the case of both roots this would map one element in the domain to two elements in the range. This is sometimes called a one to many relationship.

See graph: