Graph y^2+4<=x^2

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

Step 1

Subtract $4$ from both sides of the inequality.

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

Step 2

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

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

Step 3

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

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

Simplify the right side.

Simplify $\sqrt{x^{2} - 4}$ .

Rewrite $4$ as $2^{2}$ .

$| y | \leq \sqrt{x^{2} - 2^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$| y | \leq \sqrt{(x + 2) (x - 2)}$

Step 4

Write $| y | \leq \sqrt{(x + 2) (x - 2)}$ 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.

$y \leq \sqrt{(x + 2) (x - 2)}$

Find the domain of $y \leq \sqrt{(x + 2) (x - 2)}$ and find the intersection with $y \geq 0$ .

Find the domain of $y \leq \sqrt{(x + 2) (x - 2)}$ .

Set the radicand in $\sqrt{(x + 2) (x - 2)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 2) (x - 2) \geq 0$

Solve for $y$ .

Simplify $(x + 2) (x - 2)$ .

Expand $(x + 2) (x - 2)$ using the FOIL Method.

Apply the distributive property.

$x (x - 2) + 2 (x - 2) \geq 0$

Apply the distributive property.

$x \cdot x + x \cdot - 2 + 2 (x - 2) \geq 0$

Apply the distributive property.

$x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2 \geq 0$

Simplify terms.

Combine the opposite terms in $x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2$ .

Reorder the factors in the terms $x \cdot - 2$ and $2 x$ .

$x \cdot x - 2 x + 2 x + 2 \cdot - 2 \geq 0$

Add $- 2 x$ and $2 x$ .

$x \cdot x + 0 + 2 \cdot - 2 \geq 0$

Add $x \cdot x$ and $0$ .

$x \cdot x + 2 \cdot - 2 \geq 0$

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + 2 \cdot - 2 \geq 0$

Multiply $2$ by $- 2$ .

$x^{2} - 4 \geq 0$

Add $4$ to both sides of the inequality.

$x^{2} \geq 4$

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

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

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

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

Simplify the right side.

Simplify $\sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$| x | \geq \sqrt{2^{2}}$

Pull terms out from under the radical.

$| x | \geq | 2 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$| x | \geq 2$

Write $| x | \geq 2$ as a piecewise.

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

$x \geq 0$

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

$x \geq 2$

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

$x < 0$

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

$- x \geq 2$

Write as a piecewise.

${\begin{cases} x \geq 2 & x \geq 0 \\ - x \geq 2 & x < 0 \end{cases}$

Find the intersection of $x \geq 2$ and $x \geq 0$ .

$x \geq 2$

Divide each term in $- x \geq 2$ by $- 1$ and simplify.

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

$\frac{- x}{- 1} \leq \frac{2}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{2}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{2}{- 1}$

Simplify the right side.

Divide $2$ by $- 1$ .

$x \leq - 2$

Find the union of the solutions.

$x \leq - 2$ or $x \geq 2$

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

$(- \infty, - 2] \cup [2, \infty)$

Find the intersection of $y \geq 0$ and $(- \infty, - 2] \cup [2, \infty)$ .

$y \geq 2$

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 $- 1$ .

$- y \leq \sqrt{(x + 2) (x - 2)}$

Find the domain of $- y \leq \sqrt{(x + 2) (x - 2)}$ and find the intersection with $y < 0$ .

Find the domain of $- y \leq \sqrt{(x + 2) (x - 2)}$ .

Set the radicand in $\sqrt{(x + 2) (x - 2)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 2) (x - 2) \geq 0$

Solve for $y$ .

Simplify $(x + 2) (x - 2)$ .

Expand $(x + 2) (x - 2)$ using the FOIL Method.

Apply the distributive property.

$x (x - 2) + 2 (x - 2) \geq 0$

Apply the distributive property.

$x \cdot x + x \cdot - 2 + 2 (x - 2) \geq 0$

Apply the distributive property.

$x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2 \geq 0$

Simplify terms.

Combine the opposite terms in $x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2$ .

Reorder the factors in the terms $x \cdot - 2$ and $2 x$ .

$x \cdot x - 2 x + 2 x + 2 \cdot - 2 \geq 0$

Add $- 2 x$ and $2 x$ .

$x \cdot x + 0 + 2 \cdot - 2 \geq 0$

Add $x \cdot x$ and $0$ .

$x \cdot x + 2 \cdot - 2 \geq 0$

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + 2 \cdot - 2 \geq 0$

Multiply $2$ by $- 2$ .

$x^{2} - 4 \geq 0$

Add $4$ to both sides of the inequality.

$x^{2} \geq 4$

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

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

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

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

Simplify the right side.

Simplify $\sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$| x | \geq \sqrt{2^{2}}$

Pull terms out from under the radical.

$| x | \geq | 2 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$| x | \geq 2$

Write $| x | \geq 2$ as a piecewise.

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

$x \geq 0$

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

$x \geq 2$

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

$x < 0$

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

$- x \geq 2$

Write as a piecewise.

${\begin{cases} x \geq 2 & x \geq 0 \\ - x \geq 2 & x < 0 \end{cases}$

Find the intersection of $x \geq 2$ and $x \geq 0$ .

$x \geq 2$

Divide each term in $- x \geq 2$ by $- 1$ and simplify.

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

$\frac{- x}{- 1} \leq \frac{2}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{2}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{2}{- 1}$

Simplify the right side.

Divide $2$ by $- 1$ .

$x \leq - 2$

Find the union of the solutions.

$x \leq - 2$ or $x \geq 2$

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

$(- \infty, - 2] \cup [2, \infty)$

Find the intersection of $y < 0$ and $(- \infty, - 2] \cup [2, \infty)$ .

$y \leq - 2$

Write as a piecewise.

${\begin{cases} y \leq \sqrt{(x + 2) (x - 2)} & y \geq 2 \\ - y \leq \sqrt{(x + 2) (x - 2)} & y \leq - 2 \end{cases}$

Step 5

Find the intersection of $y \leq \sqrt{(x + 2) (x - 2)}$ and $y \geq 2$ .

$2 \leq y \leq \sqrt{(x + 2) (x - 2)}$

Step 6

Solve $- y \leq \sqrt{(x + 2) (x - 2)}$ when $y \leq - 2$ .

Divide each term in $- y \leq \sqrt{(x + 2) (x - 2)}$ by $- 1$ and simplify.

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

$\frac{- y}{- 1} \geq \frac{\sqrt{(x + 2) (x - 2)}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} \geq \frac{\sqrt{(x + 2) (x - 2)}}{- 1}$

Divide $y$ by $1$ .

$y \geq \frac{\sqrt{(x + 2) (x - 2)}}{- 1}$

Simplify the right side.

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

$y \geq - 1 \cdot \sqrt{(x + 2) (x - 2)}$

Rewrite $- 1 \cdot \sqrt{(x + 2) (x - 2)}$ as $- \sqrt{(x + 2) (x - 2)}$ .

$y \geq - \sqrt{(x + 2) (x - 2)}$

Find the intersection of $y \geq - \sqrt{(x + 2) (x - 2)}$ and $y \leq - 2$ .

$- \sqrt{(x + 2) (x - 2)} \leq y \leq - 2$

Step 7

Find the union of the solutions.

$2 \leq y \leq \sqrt{(x + 2) (x - 2)}$ or $- \sqrt{(x + 2) (x - 2)} \leq y \leq - 2$

Step 8

Graph y^2+4<=x^2

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