Graph 16x^2+4y^2<=64 y>=-x^2+2

$16 x^{2} + 4 y^{2} \leq 64$ $y \geq - x^{2} + 2$

Step 1

Solve for $y$ .

Subtract $16 x^{2}$ from both sides of the inequality.

$4 y^{2} \leq 64 - 16 x^{2}$

Divide each term in $4 y^{2} \leq 64 - 16 x^{2}$ by $4$ and simplify.

Divide each term in $4 y^{2} \leq 64 - 16 x^{2}$ by $4$ .

$\frac{4 y^{2}}{4} \leq \frac{64}{4} + \frac{- 16 x^{2}}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y^{2}}{4} \leq \frac{64}{4} + \frac{- 16 x^{2}}{4}$

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

$y^{2} \leq \frac{64}{4} + \frac{- 16 x^{2}}{4}$

Simplify the right side.

Simplify each term.

Divide $64$ by $4$ .

$y^{2} \leq 16 + \frac{- 16 x^{2}}{4}$

Cancel the common factor of $- 16$ and $4$ .

Factor $4$ out of $- 16 x^{2}$ .

$y^{2} \leq 16 + \frac{4 (- 4 x^{2})}{4}$

Cancel the common factors.

Factor $4$ out of $4$ .

$y^{2} \leq 16 + \frac{4 (- 4 x^{2})}{4 (1)}$

Cancel the common factor.

$y^{2} \leq 16 + \frac{4 (- 4 x^{2})}{4 \cdot 1}$

Rewrite the expression.

$y^{2} \leq 16 + \frac{- 4 x^{2}}{1}$

Divide $- 4 x^{2}$ by $1$ .

$y^{2} \leq 16 - 4 x^{2}$

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

$\sqrt{y^{2}} \leq \sqrt{16 - 4 x^{2}}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| y | \leq \sqrt{16 - 4 x^{2}}$

Simplify the right side.

Simplify $\sqrt{16 - 4 x^{2}}$ .

Factor $4$ out of $16 - 4 x^{2}$ .

Factor $4$ out of $16$ .

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

Factor $4$ out of $- 4 x^{2}$ .

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

Factor $4$ out of $4 (4) + 4 (- x^{2})$ .

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

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

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

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

$| y | \leq \sqrt{4 (2 + x) (2 - x)}$

Rewrite $4 (2 + x) (2 - x)$ as $2^{2} ((2 + x) (2 - x))$ .

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

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

Add parentheses.

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

Pull terms out from under the radical.

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

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

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

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

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

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

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

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

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

Solve for $y$ .

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $2$ by $2$ .

$4 + 2 (- x) + x \cdot 2 + x (- x) \geq 0$

Multiply $- 1$ by $2$ .

$4 - 2 x + x \cdot 2 + x (- x) \geq 0$

Move $2$ to the left of $x$ .

$4 - 2 x + 2 \cdot x + x (- x) \geq 0$

Rewrite using the commutative property of multiplication.

$4 - 2 x + 2 x - x \cdot x \geq 0$

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

Move $x$ .

$4 - 2 x + 2 x - (x \cdot x) \geq 0$

Multiply $x$ by $x$ .

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

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

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

Add $4$ and $0$ .

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

Subtract $4$ from both sides of the inequality.

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

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

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

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

Simplify the right side.

Divide $- 4$ by $- 1$ .

$x^{2} \leq 4$

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

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

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

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

Simplify the right side.

Simplify $\sqrt{4}$ .

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

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

Pull terms out from under the radical.

$| x | \leq | 2 |$

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

$| x | \leq 2$

Write $| x | \leq 2$ 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.

$x \leq 2$

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 $- 1$ .

$- x \leq 2$

Write as a piecewise.

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

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

$0 \leq x \leq 2$

Solve $- x \leq 2$ when $x < 0$ .

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $2$ by $- 1$ .

$x \geq - 2$

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

$- 2 \leq x < 0$

Find the union of the solutions.

$- 2 \leq x \leq 2$

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

$[- 2, 2]$

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

$0 \leq y \leq 2$

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 $- 1$ .

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

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

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

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

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

Solve for $y$ .

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $2$ by $2$ .

$4 + 2 (- x) + x \cdot 2 + x (- x) \geq 0$

Multiply $- 1$ by $2$ .

$4 - 2 x + x \cdot 2 + x (- x) \geq 0$

Move $2$ to the left of $x$ .

$4 - 2 x + 2 \cdot x + x (- x) \geq 0$

Rewrite using the commutative property of multiplication.

$4 - 2 x + 2 x - x \cdot x \geq 0$

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

Move $x$ .

$4 - 2 x + 2 x - (x \cdot x) \geq 0$

Multiply $x$ by $x$ .

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

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

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

Add $4$ and $0$ .

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

Subtract $4$ from both sides of the inequality.

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

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

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

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

Simplify the right side.

Divide $- 4$ by $- 1$ .

$x^{2} \leq 4$

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

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

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

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

Simplify the right side.

Simplify $\sqrt{4}$ .

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

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

Pull terms out from under the radical.

$| x | \leq | 2 |$

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

$| x | \leq 2$

Write $| x | \leq 2$ 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.

$x \leq 2$

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 $- 1$ .

$- x \leq 2$

Write as a piecewise.

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

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

$0 \leq x \leq 2$

Solve $- x \leq 2$ when $x < 0$ .

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $2$ by $- 1$ .

$x \geq - 2$

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

$- 2 \leq x < 0$

Find the union of the solutions.

$- 2 \leq x \leq 2$

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

$[- 2, 2]$

Find the intersection of $y < 0$ and $[- 2, 2]$ .

$- 2 \leq y < 0$

Write as a piecewise.

${\begin{cases} y \leq 2 \sqrt{(2 + x) (2 - x)} & 0 \leq y \leq 2 \\ - y \leq 2 \sqrt{(2 + x) (2 - x)} & - 2 \leq y < 0 \end{cases}$

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

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

Solve $- y \leq 2 \sqrt{(2 + x) (2 - x)}$ when $- 2 \leq y < 0$ .

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $y$ by $1$ .

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

Simplify the right side.

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

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

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

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

Multiply $2$ by $- 1$ .

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

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

No solution

Find the union of the solutions.

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

Step 2

Graph $y \geq - x^{2} + 2$ .

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

Not Linear

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

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

Step 3

Plot each graph on the same coordinate system.

$16 x^{2} + 4 y^{2} \leq 64$

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

Step 4

Graph 16x^2+4y^2<=64 y>=-x^2+2

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