Graph yy<=-1/2x+1

$y \cdot y \leq - \frac{1}{2} x + 1$

Step 1

Multiply $y$ by $y$ .

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

Step 2

Combine $x$ and $\frac{1}{2}$ .

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

Step 3

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

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

Step 4

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

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

Simplify the right side.

Simplify $\sqrt{- \frac{x}{2} + 1}$ .

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Rewrite $\sqrt{\frac{- x + 2}{2}}$ as $\frac{\sqrt{- x + 2}}{\sqrt{2}}$ .

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

Multiply $\frac{\sqrt{- x + 2}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$| y | \leq \frac{\sqrt{- x + 2}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{- x + 2}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

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

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Combine $\frac{1}{2}$ and $2$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Combine using the product rule for radicals.

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

Reorder factors in $\frac{\sqrt{(- x + 2) \cdot 2}}{2}$ .

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

Step 5

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

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

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

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

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

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

Solve for $y$ .

Divide each term in $2 (- x + 2) \geq 0$ by $2$ and simplify.

Divide each term in $2 (- x + 2) \geq 0$ by $2$ .

$\frac{2 (- x + 2)}{2} \geq \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 (- x + 2)}{2} \geq \frac{0}{2}$

Divide $- x + 2$ by $1$ .

$- x + 2 \geq \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$- x + 2 \geq 0$

Subtract $2$ from both sides of the inequality.

$- x \geq - 2$

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 $- 1$ .

$x \leq 2$

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

$(- \infty, 2]$

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

$0 \leq y \leq 2$

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 $- 1$ .

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

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

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

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

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

Solve for $y$ .

Divide each term in $2 (- x + 2) \geq 0$ by $2$ and simplify.

Divide each term in $2 (- x + 2) \geq 0$ by $2$ .

$\frac{2 (- x + 2)}{2} \geq \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 (- x + 2)}{2} \geq \frac{0}{2}$

Divide $- x + 2$ by $1$ .

$- x + 2 \geq \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$- x + 2 \geq 0$

Subtract $2$ from both sides of the inequality.

$- x \geq - 2$

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 $- 1$ .

$x \leq 2$

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

$(- \infty, 2]$

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

$y < 0$

Write as a piecewise.

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

Step 6

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

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

Step 7

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

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $y$ by $1$ .

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

Simplify the right side.

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

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

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

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

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

$- \frac{\sqrt{2 (- x + 2)}}{2} \leq y < 0$

Step 8

Find the union of the solutions.

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

Step 9

Graph yy<=-1/2x+1

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