# How do you graph y=sqrt(x+1), compare it to the parent graph and what is the domain and range?

Jan 7, 2018

Domain: $\left\{x | x \ge - 1\right\}$
Range: $y \in {\mathbb{R}}_{\ge 0}$

#### Explanation:

The parent graph is $\sqrt{x}$, and $\sqrt{x + 1}$ is the same graph but moved $1$ unit to the left.

The domain of the function is when the bit inside the square root is greater than or equal to zero (otherwise it wouldn't be defined in terms of real numbers). We can find out when this is the case by solving the following inequality:
$x + 1 \ge 0$

$x + \cancel{1 - 1} \ge 0 - 1$

$x \ge - 1$

So, our domain is $\left\{x | x \ge - 1\right\}$

The range of the parent graph, $\sqrt{x}$, is all real positive numbers and zero, and a move to the left does nothing to change that, so the range of this function is also all the positive real numbers and zero, ${\mathbb{R}}_{\ge 0}$.

Jan 7, 2018

Quadratic of form $\subset$

${x}_{\text{intercept}} \to \left(x , y\right) = \left(- 1 , 0\right)$
${y}_{\text{intercepts}} \to \left(x , y\right) = \left(0 , - 1\right) \mathmr{and} \left(0 , + 1\right)$

Domain ('input for $x$) $\to x \ge - 1$
Range ('output' for $y$) $\to y \ge \pm \infty$

#### Explanation:

This is a quadratic in $y$ instead of the normally seen quadratic in $x$

It is the same as the form $\cup$ rotated about the origin ${90}^{o}$ clockwise. So it ends up as form $\subset$

Square both sides

${y}^{2} = x + 1$

$x = {y}^{2} - 1$

Consider the standardised form of $x = {y}^{2} + b y + c$

$b$ has the value of 0 so the graph is symmetrical about the x-axis.

${x}_{\text{intercept}} = c = - 1$

${y}_{\text{intercepts}}$ are at $x = 0 \implies {y}^{2} - 1$

${y}_{\text{intercepts}} \to y = \pm 1$