Find the Domain and Range y^2=x+1

$y^{2} = x + 1$

Step 1

Take the square root of both sides of the equation to eliminate the exponent on the left side.

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

Step 2

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y = \sqrt{x + 1}$

Next, use the negative value of the $\pm$ to find the second solution.

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

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt{x + 1}$

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

$y = \sqrt{x + 1}$

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

Step 3

Set the radicand in $\sqrt{x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$x + 1 \geq 0$

Step 4

Subtract $1$ from both sides of the inequality.

$x \geq - 1$

Step 5

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$[- 1, \infty)$

Set-Builder Notation:

${x | x \geq - 1}$

Step 6

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${y | y \in ℝ}$

Step 7

Determine the domain and range.

Domain: $[- 1, \infty), {x | x \geq - 1}$

Range: $(- \infty, \infty), {y | y \in ℝ}$

Step 8