Graph 16(x-3)^2+4y^2<=64 y<=-|x-2|+2

$16 {(x - 3)}^{2} + 4 y^{2} \leq 64$ $y \leq - | x - 2 | + 2$

Step 1

Solve for $y$ .

Subtract $16 {(x - 3)}^{2}$ from both sides of the inequality.

$4 y^{2} \leq 64 - 16 {(x - 3)}^{2}$

Divide each term in $4 y^{2} \leq 64 - 16 {(x - 3)}^{2}$ by $4$ and simplify.

Divide each term in $4 y^{2} \leq 64 - 16 {(x - 3)}^{2}$ by $4$ .

$\frac{4 y^{2}}{4} \leq \frac{64}{4} + \frac{- 16 {(x - 3)}^{2}}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y^{2}}{4} \leq \frac{64}{4} + \frac{- 16 {(x - 3)}^{2}}{4}$

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

$y^{2} \leq \frac{64}{4} + \frac{- 16 {(x - 3)}^{2}}{4}$

Simplify the right side.

Simplify each term.

Divide $64$ by $4$ .

$y^{2} \leq 16 + \frac{- 16 {(x - 3)}^{2}}{4}$

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

Factor $4$ out of $- 16 {(x - 3)}^{2}$ .

$y^{2} \leq 16 + \frac{4 (- 4 {(x - 3)}^{2})}{4}$

Cancel the common factors.

Factor $4$ out of $4$ .

$y^{2} \leq 16 + \frac{4 (- 4 {(x - 3)}^{2})}{4 (1)}$

Cancel the common factor.

$y^{2} \leq 16 + \frac{4 (- 4 {(x - 3)}^{2})}{4 \cdot 1}$

Rewrite the expression.

$y^{2} \leq 16 + \frac{- 4 {(x - 3)}^{2}}{1}$

Divide $- 4 {(x - 3)}^{2}$ by $1$ .

$y^{2} \leq 16 - 4 {(x - 3)}^{2}$

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

$\sqrt{y^{2}} \leq \sqrt{16 - 4 {(x - 3)}^{2}}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| y | \leq \sqrt{16 - 4 {(x - 3)}^{2}}$

Simplify the right side.

Simplify $\sqrt{16 - 4 {(x - 3)}^{2}}$ .

Simplify with factoring out.

Factor $4$ out of $16 - 4 {(x - 3)}^{2}$ .

Factor $4$ out of $16$ .

$| y | \leq \sqrt{4 (4) - 4 {(x - 3)}^{2}}$

Factor $4$ out of $- 4 {(x - 3)}^{2}$ .

$| y | \leq \sqrt{4 (4) + 4 (- {(x - 3)}^{2})}$

Factor $4$ out of $4 (4) + 4 (- {(x - 3)}^{2})$ .

$| y | \leq \sqrt{4 (4 - {(x - 3)}^{2})}$

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

$| y | \leq \sqrt{4 (2^{2} - {(x - 3)}^{2})}$

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 $b = x - 3$ .

$| y | \leq \sqrt{4 ((2 + x - 3) (2 - (x - 3)))}$

Simplify.

Subtract $3$ from $2$ .

$| y | \leq \sqrt{4 ((x - 1) (2 - (x - 3)))}$

Apply the distributive property.

$| y | \leq \sqrt{4 ((x - 1) (2 - x - - 3))}$

Multiply $- 1$ by $- 3$ .

$| y | \leq \sqrt{4 ((x - 1) (2 - x + 3))}$

Add $2$ and $3$ .

$| y | \leq \sqrt{4 ((x - 1) (- x + 5))}$

$| y | \leq \sqrt{4 (x - 1) (- x + 5)}$

Rewrite $4 (x - 1) (- x + 5)$ as $2^{2} ((x - 1) (- x + 5))$ .

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

$| y | \leq \sqrt{2^{2} (x - 1) (- x + 5)}$

Add parentheses.

$| y | \leq \sqrt{2^{2} ((x - 1) (- x + 5))}$

Pull terms out from under the radical.

$| y | \leq | 2 | \sqrt{(x - 1) (- x + 5)}$

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

$| y | \leq 2 \sqrt{(x - 1) (- x + 5)}$

Write $| y | \leq 2 \sqrt{(x - 1) (- x + 5)}$ 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.

$y \leq 2 \sqrt{(x - 1) (- x + 5)}$

Find the domain of $y \leq 2 \sqrt{(x - 1) (- x + 5)}$ and find the intersection with $y \geq 0$ .

Find the domain of $y \leq 2 \sqrt{(x - 1) (- x + 5)}$ .

Set the radicand in $\sqrt{(x - 1) (- x + 5)}$ greater than or equal to $0$ to find where the expression is defined.

$(x - 1) (- x + 5) \geq 0$

Solve for $y$ .

Simplify $(x - 1) (- x + 5)$ .

Expand $(x - 1) (- x + 5)$ using the FOIL Method.

Apply the distributive property.

$x (- x + 5) - 1 (- x + 5) \geq 0$

Apply the distributive property.

$x (- x) + x \cdot 5 - 1 (- x + 5) \geq 0$

Apply the distributive property.

$x (- x) + x \cdot 5 - 1 (- x) - 1 \cdot 5 \geq 0$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$- x \cdot x + x \cdot 5 - 1 (- x) - 1 \cdot 5 \geq 0$

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

Move $x$ .

$- (x \cdot x) + x \cdot 5 - 1 (- x) - 1 \cdot 5 \geq 0$

Multiply $x$ by $x$ .

$- x^{2} + x \cdot 5 - 1 (- x) - 1 \cdot 5 \geq 0$

Move $5$ to the left of $x$ .

$- x^{2} + 5 \cdot x - 1 (- x) - 1 \cdot 5 \geq 0$

Multiply $- 1 (- x)$ .

Multiply $- 1$ by $- 1$ .

$- x^{2} + 5 x + 1 x - 1 \cdot 5 \geq 0$

Multiply $x$ by $1$ .

$- x^{2} + 5 x + x - 1 \cdot 5 \geq 0$

Multiply $- 1$ by $5$ .

$- x^{2} + 5 x + x - 5 \geq 0$

Add $5 x$ and $x$ .

$- x^{2} + 6 x - 5 \geq 0$

Convert the inequality to an equation.

$- x^{2} + 6 x - 5 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 6 x - 5$ .

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 6 x - 5 = 0$

Factor $- 1$ out of $6 x$ .

$- (x^{2}) - (- 6 x) - 5 = 0$

Rewrite $- 5$ as $- 1 (5)$ .

$- (x^{2}) - (- 6 x) - 1 \cdot 5 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 6 x)$ .

$- (x^{2} - 6 x) - 1 \cdot 5 = 0$

Factor $- 1$ out of $- (x^{2} - 6 x) - 1 (5)$ .

$- (x^{2} - 6 x + 5) = 0$

Factor.

Factor $x^{2} - 6 x + 5$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $5$ and whose sum is $- 6$ .

$- 5, - 1$

Write the factored form using these integers.

$- ((x - 5) (x - 1)) = 0$

Remove unnecessary parentheses.

$- (x - 5) (x - 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 5 = 0$

$x - 1 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $- (x - 5) (x - 1) = 0$ true.

$x = 5, 1$

Use each root to create test intervals.

$x < 1$

$1 < x < 5$

$x > 5$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < 1$ to see if it makes the inequality true.

Choose a value on the interval $x < 1$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$((0) - 1) (- (0) + 5) \geq 0$

The left side $- 5$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $1 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 5$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

$((3) - 1) (- (3) + 5) \geq 0$

The left side $4$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 5$ to see if it makes the inequality true.

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

$((8) - 1) (- (8) + 5) \geq 0$

The left side $- 21$ is less than the right side $0$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < 1$ False

$1 < x < 5$ True

$x > 5$ False

$x < 1$ False

$1 < x < 5$ True

$x > 5$ False

The solution consists of all of the true intervals.

$1 \leq x \leq 5$

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

$[1, 5]$

Find the intersection of $y \geq 0$ and $[1, 5]$ .

$1 \leq y \leq 5$

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 $- 1$ .

$- y \leq 2 \sqrt{(x - 1) (- x + 5)}$

Find the domain of $- y \leq 2 \sqrt{(x - 1) (- x + 5)}$ and find the intersection with $y < 0$ .

Find the domain of $- y \leq 2 \sqrt{(x - 1) (- x + 5)}$ .

Set the radicand in $\sqrt{(x - 1) (- x + 5)}$ greater than or equal to $0$ to find where the expression is defined.

$(x - 1) (- x + 5) \geq 0$

Solve for $y$ .

Simplify $(x - 1) (- x + 5)$ .

Expand $(x - 1) (- x + 5)$ using the FOIL Method.

Apply the distributive property.

$x (- x + 5) - 1 (- x + 5) \geq 0$

Apply the distributive property.

$x (- x) + x \cdot 5 - 1 (- x + 5) \geq 0$

Apply the distributive property.

$x (- x) + x \cdot 5 - 1 (- x) - 1 \cdot 5 \geq 0$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$- x \cdot x + x \cdot 5 - 1 (- x) - 1 \cdot 5 \geq 0$

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

Move $x$ .

$- (x \cdot x) + x \cdot 5 - 1 (- x) - 1 \cdot 5 \geq 0$

Multiply $x$ by $x$ .

$- x^{2} + x \cdot 5 - 1 (- x) - 1 \cdot 5 \geq 0$

Move $5$ to the left of $x$ .

$- x^{2} + 5 \cdot x - 1 (- x) - 1 \cdot 5 \geq 0$

Multiply $- 1 (- x)$ .

Multiply $- 1$ by $- 1$ .

$- x^{2} + 5 x + 1 x - 1 \cdot 5 \geq 0$

Multiply $x$ by $1$ .

$- x^{2} + 5 x + x - 1 \cdot 5 \geq 0$

Multiply $- 1$ by $5$ .

$- x^{2} + 5 x + x - 5 \geq 0$

Add $5 x$ and $x$ .

$- x^{2} + 6 x - 5 \geq 0$

Convert the inequality to an equation.

$- x^{2} + 6 x - 5 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 6 x - 5$ .

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 6 x - 5 = 0$

Factor $- 1$ out of $6 x$ .

$- (x^{2}) - (- 6 x) - 5 = 0$

Rewrite $- 5$ as $- 1 (5)$ .

$- (x^{2}) - (- 6 x) - 1 \cdot 5 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 6 x)$ .

$- (x^{2} - 6 x) - 1 \cdot 5 = 0$

Factor $- 1$ out of $- (x^{2} - 6 x) - 1 (5)$ .

$- (x^{2} - 6 x + 5) = 0$

Factor.

Factor $x^{2} - 6 x + 5$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $5$ and whose sum is $- 6$ .

$- 5, - 1$

Write the factored form using these integers.

$- ((x - 5) (x - 1)) = 0$

Remove unnecessary parentheses.

$- (x - 5) (x - 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 5 = 0$

$x - 1 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $- (x - 5) (x - 1) = 0$ true.

$x = 5, 1$

Use each root to create test intervals.

$x < 1$

$1 < x < 5$

$x > 5$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < 1$ to see if it makes the inequality true.

Choose a value on the interval $x < 1$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$((0) - 1) (- (0) + 5) \geq 0$

The left side $- 5$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $1 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 5$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

$((3) - 1) (- (3) + 5) \geq 0$

The left side $4$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 5$ to see if it makes the inequality true.

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

$((8) - 1) (- (8) + 5) \geq 0$

The left side $- 21$ is less than the right side $0$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < 1$ False

$1 < x < 5$ True

$x > 5$ False

$x < 1$ False

$1 < x < 5$ True

$x > 5$ False

The solution consists of all of the true intervals.

$1 \leq x \leq 5$

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

$[1, 5]$

Find the intersection of $y < 0$ and $[1, 5]$ .

$No solution$

Write as a piecewise.

${\begin{cases} y \leq 2 \sqrt{(x - 1) (- x + 5)} & 1 \leq y \leq 5 \end{cases}$

Find the intersection of $y \leq 2 \sqrt{(x - 1) (- x + 5)}$ and $1 \leq y \leq 5$ .

$1 \leq y \leq N o (Min i m u m)$

Find the union of the solutions.

$1 \leq y \leq i N o Min m^{2} u$

Step 2

Graph $y \leq - | x - 2 | + 2$ .

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

Not Linear

Graph a solid line, then shade the area below the boundary line since $y$ is less than $- | x - 2 | + 2$ .

$y \leq - | x - 2 | + 2$

Step 3

Plot each graph on the same coordinate system.

$16 {(x - 3)}^{2} + 4 y^{2} \leq 64$

$y \leq - | x - 2 | + 2$

Step 4

Graph 16(x-3)^2+4y^2<=64 y<=-|x-2|+2

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