Graph yy<-3/2x-1

$y \cdot y < - \frac{3}{2} x - 1$

Step 1

Solve for $y$ .

Multiply $y$ by $y$ .

$y^{2} < - \frac{3}{2} x - 1$

Simplify each term.

Combine $x$ and $\frac{3}{2}$ .

$y^{2} < - \frac{x \cdot 3}{2} - 1$

Move $3$ to the left of $x$ .

$y^{2} < - \frac{3 x}{2} - 1$

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

$\sqrt{y^{2}} < \sqrt{- \frac{3 x}{2} - 1}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| y | < \sqrt{- \frac{3 x}{2} - 1}$

Simplify the right side.

Simplify $\sqrt{- \frac{3 x}{2} - 1}$ .

Factor $- 1$ out of $- \frac{3 x}{2} - 1$ .

Factor $- 1$ out of $- \frac{3 x}{2}$ .

$| y | < \sqrt{- (\frac{3 x}{2}) - 1}$

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

$| y | < \sqrt{- (\frac{3 x}{2}) - 1 (1)}$

Factor $- 1$ out of $- (\frac{3 x}{2}) - 1 (1)$ .

$| y | < \sqrt{- (\frac{3 x}{2} + 1)}$

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

$| y | < \sqrt{- (\frac{3 x}{2} + \frac{2}{2})}$

Combine the numerators over the common denominator.

$| y | < \sqrt{- \frac{3 x + 2}{2}}$

Write $| y | < \sqrt{- \frac{3 x + 2}{2}}$ 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.

$y < \sqrt{- \frac{3 x + 2}{2}}$

Find the domain of $y < \sqrt{- \frac{3 x + 2}{2}}$ and find the intersection with $y \geq 0$ .

Find the domain of $y < \sqrt{- \frac{3 x + 2}{2}}$ .

Set the radicand in $\sqrt{- \frac{3 x + 2}{2}}$ greater than or equal to $0$ to find where the expression is defined.

$- \frac{3 x + 2}{2} \geq 0$

Solve for $y$ .

Divide each term in $- \frac{3 x + 2}{2} \geq 0$ by $- 1$ and simplify.

Divide each term in $- \frac{3 x + 2}{2} \geq 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- \frac{3 x + 2}{2}}{- 1} \leq \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\frac{3 x + 2}{2}}{1} \leq \frac{0}{- 1}$

Divide $\frac{3 x + 2}{2}$ by $1$ .

$\frac{3 x + 2}{2} \leq \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

$\frac{3 x + 2}{2} \leq 0$

Multiply both sides by $2$ .

$\frac{3 x + 2}{2} \cdot 2 \leq 0 \cdot 2$

Simplify.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{3 x + 2}{2} \cdot 2 \leq 0 \cdot 2$

Rewrite the expression.

$3 x + 2 \leq 0 \cdot 2$

Simplify the right side.

Multiply $0$ by $2$ .

$3 x + 2 \leq 0$

Solve for $x$ .

Subtract $2$ from both sides of the inequality.

$3 x \leq - 2$

Divide each term in $3 x \leq - 2$ by $3$ and simplify.

Divide each term in $3 x \leq - 2$ by $3$ .

$\frac{3 x}{3} \leq \frac{- 2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \leq \frac{- 2}{3}$

Divide $x$ by $1$ .

$x \leq \frac{- 2}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x \leq - \frac{2}{3}$

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

$(- \infty, - \frac{2}{3}]$

Find the intersection of $y \geq 0$ and $(- \infty, - \frac{2}{3}]$ .

$No solution$

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 $- 1$ .

$- y < \sqrt{- \frac{3 x + 2}{2}}$

Find the domain of $- y < \sqrt{- \frac{3 x + 2}{2}}$ and find the intersection with $y < 0$ .

Find the domain of $- y < \sqrt{- \frac{3 x + 2}{2}}$ .

Set the radicand in $\sqrt{- \frac{3 x + 2}{2}}$ greater than or equal to $0$ to find where the expression is defined.

$- \frac{3 x + 2}{2} \geq 0$

Solve for $y$ .

Divide each term in $- \frac{3 x + 2}{2} \geq 0$ by $- 1$ and simplify.

Divide each term in $- \frac{3 x + 2}{2} \geq 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- \frac{3 x + 2}{2}}{- 1} \leq \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\frac{3 x + 2}{2}}{1} \leq \frac{0}{- 1}$

Divide $\frac{3 x + 2}{2}$ by $1$ .

$\frac{3 x + 2}{2} \leq \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

$\frac{3 x + 2}{2} \leq 0$

Multiply both sides by $2$ .

$\frac{3 x + 2}{2} \cdot 2 \leq 0 \cdot 2$

Simplify.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{3 x + 2}{2} \cdot 2 \leq 0 \cdot 2$

Rewrite the expression.

$3 x + 2 \leq 0 \cdot 2$

Simplify the right side.

Multiply $0$ by $2$ .

$3 x + 2 \leq 0$

Solve for $x$ .

Subtract $2$ from both sides of the inequality.

$3 x \leq - 2$

Divide each term in $3 x \leq - 2$ by $3$ and simplify.

Divide each term in $3 x \leq - 2$ by $3$ .

$\frac{3 x}{3} \leq \frac{- 2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \leq \frac{- 2}{3}$

Divide $x$ by $1$ .

$x \leq \frac{- 2}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x \leq - \frac{2}{3}$

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

$(- \infty, - \frac{2}{3}]$

Find the intersection of $y < 0$ and $(- \infty, - \frac{2}{3}]$ .

$y \leq - \frac{2}{3}$

Write as a piecewise.

${\begin{cases} - y < \sqrt{- \frac{3 x + 2}{2}} & y \leq - \frac{2}{3} \end{cases}$

Solve $- y < \sqrt{- \frac{3 x + 2}{2}}$ when $y \leq - \frac{2}{3}$ .

Divide each term in $- y < \sqrt{- \frac{3 x + 2}{2}}$ by $- 1$ and simplify.

Divide each term in $- y < \sqrt{- \frac{3 x + 2}{2}}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} > \frac{\sqrt{- \frac{3 x + 2}{2}}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} > \frac{\sqrt{- \frac{3 x + 2}{2}}}{- 1}$

Divide $y$ by $1$ .

$y > \frac{\sqrt{- \frac{3 x + 2}{2}}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{- \frac{3 x + 2}{2}}}{- 1}$ .

$y > - 1 \cdot \sqrt{- \frac{3 x + 2}{2}}$

Rewrite $- 1 \cdot \sqrt{- \frac{3 x + 2}{2}}$ as $- \sqrt{- \frac{3 x + 2}{2}}$ .

$y > - \sqrt{- \frac{3 x + 2}{2}}$

Find the intersection of $y > - \sqrt{- \frac{3 x + 2}{2}}$ and $y \leq - \frac{2}{3}$ .

$- \sqrt{- \frac{3 x + 2}{2}} < y \leq - \frac{2}{3}$

Find the union of the solutions.

$- \sqrt{- \frac{3 x + 2}{2}} < y \leq - \frac{2}{3}$

Step 2

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

Not Linear

Step 3

Graph a solid line, then shade the area below the boundary line since $y$ is less than .

$- \sqrt{- \frac{3 x + 2}{2}} < y \leq - \frac{2}{3}$

Step 4

Graph yy<-3/2x-1

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