Graph x^2+y^2<=16 x^2-y<2

$x^{2} + y^{2} \leq 16$ $x^{2} - y < 2$

Solve for $y$ .

Subtract $x^{2}$ from both sides of the inequality.

$y^{2} \leq 16 - x^{2}$

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

$\sqrt{y^{2}} \leq \sqrt{16 - x^{2}}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| y | \leq \sqrt{16 - x^{2}}$

Simplify the right side.

Simplify $\sqrt{16 - x^{2}}$ .

Rewrite $16$ as $4^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4$ and $b = x$ .

$| y | \leq \sqrt{(4 + x) (4 - x)}$

Write $| y | \leq \sqrt{(4 + x) (4 - x)}$ 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.

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

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

Find the domain of $y \leq \sqrt{(4 + x) (4 - x)}$ .

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

$(4 + x) (4 - x) \geq 0$

Solve for $y$ .

Simplify $(4 + x) (4 - x)$ .

Expand $(4 + x) (4 - x)$ using the FOIL Method.

Apply the distributive property.

$4 (4 - x) + x (4 - x) \geq 0$

Apply the distributive property.

$4 \cdot 4 + 4 (- x) + x (4 - x) \geq 0$

Apply the distributive property.

$4 \cdot 4 + 4 (- x) + x \cdot 4 + x (- x) \geq 0$

Simplify and combine like terms.

Simplify each term.

Multiply $4$ by $4$ .

$16 + 4 (- x) + x \cdot 4 + x (- x) \geq 0$

Multiply $- 1$ by $4$ .

$16 - 4 x + x \cdot 4 + x (- x) \geq 0$

Move $4$ to the left of $x$ .

$16 - 4 x + 4 \cdot x + x (- x) \geq 0$

Rewrite using the commutative property of multiplication.

$16 - 4 x + 4 x - x \cdot x \geq 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$16 - 4 x + 4 x - (x \cdot x) \geq 0$

Multiply $x$ by $x$ .

$16 - 4 x + 4 x - x^{2} \geq 0$

Add $- 4 x$ and $4 x$ .

$16 + 0 - x^{2} \geq 0$

Add $16$ and $0$ .

$16 - x^{2} \geq 0$

Subtract $16$ from both sides of the inequality.

$- x^{2} \geq - 16$

Divide each term in $- x^{2} \geq - 16$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 16$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 16}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 16}{- 1}$

Divide $x^{2}$ by $1$ .

$x^{2} \leq \frac{- 16}{- 1}$

Simplify the right side.

Divide $- 16$ by $- 1$ .

$x^{2} \leq 16$

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

$\sqrt{x^{2}} \leq \sqrt{16}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{16}$

Simplify the right side.

Simplify $\sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

$| x | \leq | 4 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$| x | \leq 4$

Write $| x | \leq 4$ 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.

$x \leq 4$

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 $- 1$ .

$- x \leq 4$

Write as a piecewise.

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

Find the intersection of $x \leq 4$ and $x \geq 0$ .

$0 \leq x \leq 4$

Solve $- x \leq 4$ when $x < 0$ .

Divide each term in $- x \leq 4$ by $- 1$ and simplify.

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

$\frac{- x}{- 1} \geq \frac{4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{4}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{4}{- 1}$

Simplify the right side.

Divide $4$ by $- 1$ .

$x \geq - 4$

Find the intersection of $x \geq - 4$ and $x < 0$ .

$- 4 \leq x < 0$

Find the union of the solutions.

$- 4 \leq x \leq 4$

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

$[- 4, 4]$

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

$0 \leq y \leq 4$

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 $- 1$ .

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

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

Find the domain of $- y \leq \sqrt{(4 + x) (4 - x)}$ .

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

$(4 + x) (4 - x) \geq 0$

Solve for $y$ .

Simplify $(4 + x) (4 - x)$ .

Expand $(4 + x) (4 - x)$ using the FOIL Method.

Apply the distributive property.

$4 (4 - x) + x (4 - x) \geq 0$

Apply the distributive property.

$4 \cdot 4 + 4 (- x) + x (4 - x) \geq 0$

Apply the distributive property.

$4 \cdot 4 + 4 (- x) + x \cdot 4 + x (- x) \geq 0$

Simplify and combine like terms.

Simplify each term.

Multiply $4$ by $4$ .

$16 + 4 (- x) + x \cdot 4 + x (- x) \geq 0$

Multiply $- 1$ by $4$ .

$16 - 4 x + x \cdot 4 + x (- x) \geq 0$

Move $4$ to the left of $x$ .

$16 - 4 x + 4 \cdot x + x (- x) \geq 0$

Rewrite using the commutative property of multiplication.

$16 - 4 x + 4 x - x \cdot x \geq 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$16 - 4 x + 4 x - (x \cdot x) \geq 0$

Multiply $x$ by $x$ .

$16 - 4 x + 4 x - x^{2} \geq 0$

Add $- 4 x$ and $4 x$ .

$16 + 0 - x^{2} \geq 0$

Add $16$ and $0$ .

$16 - x^{2} \geq 0$

Subtract $16$ from both sides of the inequality.

$- x^{2} \geq - 16$

Divide each term in $- x^{2} \geq - 16$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 16$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 16}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 16}{- 1}$

Divide $x^{2}$ by $1$ .

$x^{2} \leq \frac{- 16}{- 1}$

Simplify the right side.

Divide $- 16$ by $- 1$ .

$x^{2} \leq 16$

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

$\sqrt{x^{2}} \leq \sqrt{16}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{16}$

Simplify the right side.

Simplify $\sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

$| x | \leq | 4 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$| x | \leq 4$

Write $| x | \leq 4$ 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.

$x \leq 4$

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 $- 1$ .

$- x \leq 4$

Write as a piecewise.

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

Find the intersection of $x \leq 4$ and $x \geq 0$ .

$0 \leq x \leq 4$

Solve $- x \leq 4$ when $x < 0$ .

Divide each term in $- x \leq 4$ by $- 1$ and simplify.

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

$\frac{- x}{- 1} \geq \frac{4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{4}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{4}{- 1}$

Simplify the right side.

Divide $4$ by $- 1$ .

$x \geq - 4$

Find the intersection of $x \geq - 4$ and $x < 0$ .

$- 4 \leq x < 0$

Find the union of the solutions.

$- 4 \leq x \leq 4$

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

$[- 4, 4]$

Find the intersection of $y < 0$ and $[- 4, 4]$ .

$- 4 \leq y < 0$

Write as a piecewise.

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

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

$0 \leq y \leq N o (Min i m u m)$

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

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

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

$\frac{- y}{- 1} \geq \frac{\sqrt{(4 + x) (4 - x)}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} \geq \frac{\sqrt{(4 + x) (4 - x)}}{- 1}$

Divide $y$ by $1$ .

$y \geq \frac{\sqrt{(4 + x) (4 - x)}}{- 1}$

Simplify the right side.

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

$y \geq - 1 \cdot \sqrt{(4 + x) (4 - x)}$

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

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

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

No solution

Find the union of the solutions.

$0 \leq y \leq i N o Min m^{2} u$

Graph $x^{2} - y < 2$ .

Solve for $y$ .

Subtract $x^{2}$ from both sides of the inequality.

$- y < 2 - x^{2}$

Divide each term in $- y < 2 - x^{2}$ by $- 1$ and simplify.

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

$\frac{- y}{- 1} > \frac{2}{- 1} + \frac{- x^{2}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} > \frac{2}{- 1} + \frac{- x^{2}}{- 1}$

Divide $y$ by $1$ .

$y > \frac{2}{- 1} + \frac{- x^{2}}{- 1}$

Simplify the right side.

Simplify each term.

Divide $2$ by $- 1$ .

$y > - 2 + \frac{- x^{2}}{- 1}$

Dividing two negative values results in a positive value.

$y > - 2 + \frac{x^{2}}{1}$

Divide $x^{2}$ by $1$ .

$y > - 2 + x^{2}$

Find the slope and the y-intercept for the boundary line.

Rewrite in slope-intercept form.

The slope-intercept form is $y = m x + b$ , where $m$ is the slope and $b$ is the y-intercept.

$y = m x + b$

Reorder $- 2$ and $x^{2}$ .

$y > x^{2} - 2$

The equation is not linear, so a constant slope does not exist.

Not Linear

Graph a dashed line, then shade the area above the boundary line since $y$ is greater than $- 2 + x^{2}$ .

$y > - 2 + x^{2}$

Plot each graph on the same coordinate system.

$x^{2} + y^{2} \leq 16$

$x^{2} - y < 2$

Graph x^2+y^2<=16 x^2-y<2

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