Graph x = square root of y+1

$x = \sqrt{y + 1}$

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

$[- 1, \infty)$

${y | y \geq - 1}$

To find the radical expression end point, substitute the $y$ value $- 1$ , which is the least value in the domain, into $f (y) = \sqrt{y + 1}$ .

Replace the variable $y$ with $- 1$ in the expression.

$f (- 1) = \sqrt{(- 1) + 1}$

Simplify the result.

Add $- 1$ and $1$ .

$f (- 1) = \sqrt{0}$

Rewrite $0$ as $0^{2}$ .

$f (- 1) = \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$f (- 1) = 0$

The final answer is $0$ .

$0$

The radical expression end point is $(0, - 1)$ .

$(0, - 1)$

Select a few $y$ values from the domain. It would be more useful to select the values so that they are next to the $y$ value of the radical expression end point.

Substitute the $y$ value $- 1$ into $\sqrt{y + 1}$ . In this case, the point is $(0, - 1)$ .

Replace the variable $y$ with $- 1$ in the expression.

$f (- 1) = \sqrt{(- 1) + 1}$

Simplify the result.

Add $- 1$ and $1$ .

$f (- 1) = \sqrt{0}$

Rewrite $0$ as $0^{2}$ .

$f (- 1) = \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$f (- 1) = 0$

The final answer is $0$ .

$y = 0$

Substitute the $y$ value $1$ into $\sqrt{y + 1}$ . In this case, the point is $(1.41, 1)$ .

Replace the variable $y$ with $1$ in the expression.

$f (1) = \sqrt{(1) + 1}$

Simplify the result.

Add $1$ and $1$ .

$f (1) = \sqrt{2}$

The final answer is $\sqrt{2}$ .

$y = \sqrt{2}$

The square root can be graphed using the points around the vertex $(0, - 1), (0, - 1), (1.41, 1)$

$\begin{array}{c} x & y \\ 0 & - 1 \\ 0 & - 1 \\ 1.41 & 1 \end{array}$

Graph x = square root of y+1

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