Graph x>=y^2

$x \geq y^{2}$

Step 1

Rewrite so $y$ is on the left side of the inequality.

$y^{2} \leq x$

Step 2

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

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

Step 3

Simplify the left side.

Pull terms out from under the radical.

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

Step 4

Write $| y | \leq \sqrt{x}$ 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.

$y \leq \sqrt{x}$

Find the domain of $y \leq \sqrt{x}$ and find the intersection with $y \geq 0$ .

Find the domain of $y \leq \sqrt{x}$ .

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

$x \geq 0$

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

$[0, \infty)$

Find the intersection of $y \geq 0$ and $[0, \infty)$ .

$y \geq 0$

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 $- 1$ .

$- y \leq \sqrt{x}$

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

Find the domain of $- y \leq \sqrt{x}$ .

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

$x \geq 0$

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

$[0, \infty)$

Find the intersection of $y < 0$ and $[0, \infty)$ .

$No solution$

Write as a piecewise.

${\begin{cases} y \leq \sqrt{x} & y \geq 0 \end{cases}$

Step 5

Find the intersection of $y \leq \sqrt{x}$ and $y \geq 0$ .

$0 \leq y \leq \sqrt{x}$

Step 6

Find the union of the solutions.

$0 \leq y \leq \sqrt{x}$

Step 7

Graph x>=y^2

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