Graph of $\sqrt{16-x^2}$ The actual question to find the range of $\sqrt{16-x^2}$ so I think to draw Graph of $\sqrt{16-x^2}$.
 But I don't know how to draw the graph of $\sqrt{16-x^2}$ , I think graph would be a semi Circe which lies in I and II quadrant. 
Is there any another way to find the range of the given question ?
 A: The expression in question indeed would map a semicircle, centered at the origin. Generically, you can recognize the equation of a semicircle of radius $r$ and center $(h,k)$ by
$$y = k \pm \sqrt{r^2 - (x-h)^2}$$
where the positive root means its flat side is facing down, and if you take the negative root the flat side is facing up. With this in mind, you notice that the "peak" of the arc for your semicircle is right above the center $(0,0)$. Plugging in $x=0$, you would see that your equation gives $y = 4$.
Since the function in question is continuous, and you know it can only take in values from $x=-4$ to $x=4$ (so that the root isn't negative), you then test the "minimum" points, right on the edge of the diameter: $x=\pm 4$. This gives $y=0$. 
Thus you can conclude, if a little heuristically, that the range of the function is $[0,4]$. You can, alternatively, take a more algebraic approach as done in Martund's answer. Whichever you prefer.
In general, for the generic semicircle above, the range is $[k,k+r]$ for positive roots, and $[k-r,k]$ for negative roots. This makes sense since all of the points would be a distance $r$ from the center, the $y$ coordinate ($k$) of which defines a maximum or minimum for the semicircle depending on orientation.
To further convince yourself that $[0,4]$ is the range, observe the graph of the function. Desmos in general is a pretty decent graphing tool.

A: Since it is a square root, it will be non-negative. Also,
$$-x^2\leq 0$$
$$\Longrightarrow 16-x^2\leq 16$$
$$\Longrightarrow \sqrt{16-x^2}\leq 4$$
Therefore, range is $[0,4]$. Here equality holds on both ends, hence this is the required range.
Hope it helps:)
A: Make note of the following points: 


*

*Domain of $\sqrt{16-x^2}$ is the set of all points wherein $16-x^2\ge 0$ which is equivalent to saying $\mid x\mid \le 4 $.

*Maximum in the domain $\left[-4, 4\right]$ occurs when $16-x^2$ is maximum in the same domain which occurs at the point where $(16-x^2)'=0$ (Why?) which is when $x=0$. And the maximum value is $\sqrt{16-0^2}=4$.

*Minimum occurs when $16-x^2$ is minimum simultaneously satisfying $16-x^2\ge 0$, so the minimum is evidently $0$.

*By the argument of continuity we have that the range is $\left[\text{min.},\text{max.}\right]=\left[0, 4\right]$ and we are done.



Aliter:
You can also observe that the function satisfies the form that is of a circle, namely $(x-0)^2+(y-0)^2=4^2$, so it must represent a circle but since it has a "$+$" sign attached as in $+\sqrt{16-x^2}$, it must represent a semi-circle in the upper half plane with radius $4$. Why in the upper half plane, that's because "$\sqrt{\quad}$" is a positive function and always returns a positive value.
A: The graph of $y=\sqrt{16-x^2}$ is a semi circle. Fixed
A: $y=\sqrt{16-x^2}$
1)Domain: 
$D=${$x| -4 \le x \le 4$}, since the expression under the square root must be $\ge 0$.
2) Range:
Note $y \ge 0$. 
Solve for $x$:
$y^2+x^2=16.$
$x= \pm \sqrt{16 -y^2}.$
Recall $y \ge 0$.
Hence the range :
$R=${$y|$ $0\le y \le 4$}.
.
