Graph the function, and compare to the parent graph. State the domain and range.

$\mathit{y}\mathbf{=}\mathbf{-}\mathbf{2}\sqrt{\mathbf{x}\mathbf{+}\mathbf{1}}$

Parent graph:The simplest form of the given function is called the parent function of that function and the graph of the parent function is called parent graph.

Domain: The set of all possible values for which given function defined is called domain.

Range: The set of all possible values of the given function is called range.

The given function is: $y=-2\sqrt{x+1}$

In order to graph a function, find few co-ordinates by substituting values of ‘$x$’ and find finding the respective values of ‘$y$’.

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=-1\\ y=-2\sqrt{-1+1}\\ =-2\sqrt{0}\\ =-2\left(0\right)\\ =0\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=0\\ y=-2\sqrt{0+1}\\ =-2\sqrt{1}\\ =-2\left(1\right)\\ =-2\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=3,\\ y=-2\sqrt{3+1}\\ =-2\sqrt{4}\\ =-2\left(2\right)\\ =-4\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=8,\\ y=-2\sqrt{8+1}\\ =-2\sqrt{9}\\ =-2\left(3\right)\\ =-6\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=15,\\ y=-2\sqrt{15+1}\\ =-2\sqrt{16}\\ =-2\left(4\right)\\ =-8\end{array}$

Plot these co-ordinates on a coordinate plane and join those points to get the required graph.

The parent function of$y=-2\sqrt{x+1}$ is the simplest square root function.

That is,$y=\sqrt{x}$

The graph of parent function$y=\sqrt{x}$ is given below.

Note: Since the parent function is just used for comparison, it is graphed using graphing calculator.

1 is added inside square root of parent function $y=\sqrt{x}$and then is multiplied by ‘$-2$’ . So the graph $y=-2\sqrt{x+1}$is the translation and reflection of the parent graph $y=\sqrt{x}$.

Since 1 is added inside the root, the graph is translated to the left by 1 units.

Coefficient of $\sqrt{x}$is ‘$-2$’.

The absolute value of coefficient of $\sqrt{x}$is 2.

As 2 is greater than 1. The graph is a vertical stretch of the parent graph $y=\sqrt{x}$.

Also as the coefficient of $\sqrt{x}$is negative, the graph is $y=-2\sqrt{x+1}$ a reflection across the $X$-axis.

Therefore, on comparison with the parent graph, the graph $y=-2\sqrt{x+1}$ is a reflection across $x$-axis, a vertical stretch and is translated to the left by 1 units.

Since the values inside the root must be positive.

$x+1\ge 0$

Adding ‘$-1$’ on both the sides.

$\begin{array}{c}x+1-1\ge 0-1\\ x+0\ge -1\\ x\ge -1\end{array}$

Therefore,$x\ge -1\Rightarrow x\in [-1,\infty )$

Therefore, domain: $[-1,\infty )$

As the coefficient of the square root term ‘$\sqrt{x+1}$‘is ‘$-2$’ which is negative,$y$takes all the negative real values including zero.

Therefore,$y\le 0,\Rightarrow y\in (-\infty ,0]$

Therefore, Range:$(-\infty ,0]$