Graph y^2>=4+x^2

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

Step 1

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

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

Step 2

Simplify the left side.

Pull terms out from under the radical.

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

Step 3

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

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

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 $- 1$ .

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

Write as a piecewise.

${\begin{cases} y \geq \sqrt{4 + x^{2}} & y \geq 0 \\ - y \geq \sqrt{4 + x^{2}} & y < 0 \end{cases}$

Step 4

Find the intersection of $y \geq \sqrt{4 + x^{2}}$ and $y \geq 0$ .

$y \geq \sqrt{4 + x^{2}}$ and $y \geq 0$

Step 5

Solve $- y \geq \sqrt{4 + x^{2}}$ when $y < 0$ .

Divide each term in $- y \geq \sqrt{4 + x^{2}}$ by $- 1$ and simplify.

Divide each term in $- y \geq \sqrt{4 + x^{2}}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \leq \frac{\sqrt{4 + x^{2}}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} \leq \frac{\sqrt{4 + x^{2}}}{- 1}$

Divide $y$ by $1$ .

$y \leq \frac{\sqrt{4 + x^{2}}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{4 + x^{2}}}{- 1}$ .

$y \leq - 1 \cdot \sqrt{4 + x^{2}}$

Rewrite $- 1 \cdot \sqrt{4 + x^{2}}$ as $- \sqrt{4 + x^{2}}$ .

$y \leq - \sqrt{4 + x^{2}}$

Find the intersection of $y \leq - \sqrt{4 + x^{2}}$ and $y < 0$ .

$y < N o (Min i m u m)$

Step 6

Find the union of the solutions.

$y \geq N o Max i m u m$ or $y < N o Min i m u m$

Step 7

Graph y^2>=4+x^2

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