Graph x = square root of 4-y^2

$x = \sqrt{4 - y^{2}}$

Simplify $\sqrt{4 - y^{2}}$ .

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

$x = \sqrt{2^{2} - y^{2}}$

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 $b = y$ .

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

The domain in terms of $y$ are all the $y$ values that make the radicand non-negative.

$[- 2, 2]$

${y | - 2 \leq y \leq 2}$

The final answer is .

The end points are $(, S E T, x | - 2), (, 2)$ .

$(, S E T, x | - 2), (, 2)$

The square root can be graphed using the points around the vertex $(, S E T, x | - 2), (, 2),$

$\begin{array}{c} x & y \\ S E \end{array}$