Graph y^2-4>=x

$y^{2} - 4 \geq x$

Step 1

Add $4$ to both sides of the inequality.

$y^{2} \geq x + 4$

Step 2

Take the specified root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{y^{2}} \geq \sqrt{x + 4}$

Step 3

Simplify the left side.

Pull terms out from under the radical.

$| y | \geq \sqrt{x + 4}$

Step 4

Write $| y | \geq \sqrt{x + 4}$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$y \geq 0$

In the piece where $y$ is non-negative, remove the absolute value.

$y \geq \sqrt{x + 4}$

Find the domain of $y \geq \sqrt{x + 4}$ and find the intersection with $y \geq 0$ .

Find the domain of $y \geq \sqrt{x + 4}$ .

Set the radicand in $\sqrt{x + 4}$ greater than or equal to $0$ to find where the expression is defined.

$x + 4 \geq 0$

Subtract $4$ from both sides of the inequality.

$x \geq - 4$

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

$[- 4, \infty)$

Find the intersection of $y \geq 0$ and $[- 4, \infty)$ .

$y \geq 0$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$y < 0$

In the piece where $y$ is negative, remove the absolute value and multiply by $- 1$ .

$- y \geq \sqrt{x + 4}$

Find the domain of $- y \geq \sqrt{x + 4}$ and find the intersection with $y < 0$ .

Find the domain of $- y \geq \sqrt{x + 4}$ .

Set the radicand in $\sqrt{x + 4}$ greater than or equal to $0$ to find where the expression is defined.

$x + 4 \geq 0$

Subtract $4$ from both sides of the inequality.

$x \geq - 4$

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

$[- 4, \infty)$

Find the intersection of $y < 0$ and $[- 4, \infty)$ .

$- 4 \leq y < 0$

Write as a piecewise.

${\begin{cases} y \geq \sqrt{x + 4} & y \geq 0 \\ - y \geq \sqrt{x + 4} & - 4 \leq y < 0 \end{cases}$

Step 5

Find the intersection of $y \geq \sqrt{x + 4}$ and $y \geq 0$ .

$y \geq \sqrt{x + 4}$ and $y \geq 0$

Step 6

Solve $- y \geq \sqrt{x + 4}$ when $- 4 \leq y < 0$ .

Divide each term in $- y \geq \sqrt{x + 4}$ by $- 1$ and simplify.

Divide each term in $- y \geq \sqrt{x + 4}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \leq \frac{\sqrt{x + 4}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} \leq \frac{\sqrt{x + 4}}{- 1}$

Divide $y$ by $1$ .

$y \leq \frac{\sqrt{x + 4}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{x + 4}}{- 1}$ .

$y \leq - 1 \cdot \sqrt{x + 4}$

Rewrite $- 1 \cdot \sqrt{x + 4}$ as $- \sqrt{x + 4}$ .

$y \leq - \sqrt{x + 4}$

Find the intersection of $y \leq - \sqrt{x + 4}$ and $- 4 \leq y < 0$ .

$- 4 \leq y < N o (Min i m u m)$

Step 7

Find the union of the solutions.

$y \geq N o Max i m u m$ or $- 4 \leq y < i N o Min m^{2} u$

Step 8

Rewrite $y \geq N o Max i m u m or - 4 \leq y < i N o Min m^{2} u$ so $y$ is on the left side.

$\geq y \geq N o Max i m u m or - 4$

Step 9

Graph y^2-4>=x

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