Why does the graph only go to the right and up? Wouldn't there be negative values of y as well? Since, for example, the $\sqrt{4}$ is $2$ and $-2$
$\endgroup$ 26 Answers
$\begingroup$The graph of $y=\sqrt{x}$ is goes only right up since:
- there are no negative values of $x$ allowed
- there are no negative values of $y$ allowed (since $\sqrt{x}$ can never be negative)
Furthermore: $\sqrt{4}=2$, but $\sqrt{4}\ne-2$. You are confused with: $(-2)^2=2^2=4$.
$\endgroup$ 6 $\begingroup$$\sqrt{x}$ is defined only when $x\ge 0$.
For any $x\ge 0$, $\sqrt x\ge 0$. So, $\sqrt{4}=\sqrt{2^2}=2$ (not $-2).$
A function y is an output of certain operation over a variable, say x, such that it should give a unique value for any value of x. But, function can give same value for two different values of x. In a general sense, $ \sqrt{} $ is not a function, because for a single value of x, two values for y are obtained for example $$\sqrt{16} = \pm 4$$. In order to make it a function, appropriate domain for x and y are chosen such that $f:x \rightarrow y $ such that where $\sqrt{}$ is the function f with $x \in R^{+}$ and $y \in R^{+}$. Then, the graph mentioned in the question is obtained i.e curve on the right and up side of the coordinate system.
If the domain of y is defined as $ y \in R^{-} $ and domain of x remains same, then the curve would on right and downside of conventional coordinate system. If the function f maps such $\sqrt{}:x \rightarrow y$ for $x\in R^{-}$, then y have to be defined in complex plane such as to make $\sqrt{}$ as a function i.e to give unique value of y.
$\endgroup$ $\begingroup$If the same relation between x and y is taken in reverse sense i.e $g:y \rightarrow x$ such that $x = g(y)=y^2$ here squaring is the function with $y \in R$ and $x \in R^{+} $. It is the equation of parabola rotated with its concave side towards positive x-axis as shown in the following figure.
![lazy dog]()
Because $x^2$ is not one to one in real numbers. Look at the general definition of function, and it will be clear.
$\endgroup$ $\begingroup$This is a common misconception: $x^2 = 4$ means that $x=2$ or $x=-2$, HOWEVER, we never actually take the square root to find these solutions.
The square root is a "function" ($\sqrt\ :\mathbb{R^+} \rightarrow \mathbb{R})$ - and a "function" only has one input for the input you give it.
$\endgroup$ $\begingroup$A function $f$ is a map $f: X\rightarrow Y$ which for every input $x\in X$ gives precisely one output $f(x)\in Y$. Taking $\sqrt4=\pm2$ is not a function, because it has two outputs.
The notation $\sqrt{x}$ is used to denote the function $\sqrt{\:\:\:}\:: \mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ with positive image, ignoring the negative numbers which are square roots. When you graph $\sqrt{x}$, you are graphing this function.
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