#### Answer

$y \text{ is NOT a function of }x
\\\text{Domain: }
\left[ 0,\infty \right)
\\\text{NOT a linear function}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To determine if the given equation, $
x=y^2
,$ is a function, solve first for $y.$ Then check if $x$ is unique for every value of $y.$
To find the domain, find the set of all possible values of $x.$
A linear function is any equation that can be expressed as $f(x)=mx+b.$
$\bf{\text{Solution Details:}}$
Taking the square root of both sides, the given equation is equivalent to
\begin{array}{l}\require{cancel}
x=y^2
\\\\
\pm\sqrt{x}=\sqrt{y^2}
\\\\
\pm\sqrt{x}=y
\\\\
y=\pm\sqrt{x}
.\end{array}
If $x=4,$ then $y=2$ or $y=-2$. Hence, the ordered pairs, $\{(4,2),(4,-2)\}$ satisfy the given equation. Since $x$ is not unique, then $y$ is not a function of $x.$
The radicand of a radical with an even index cannot be negative. Hence,
\begin{array}{l}\require{cancel}
x\ge0
.\end{array}
The domain is the set of all nonnegative numbers.
Since the given equation cannot be expressed as $f(x)=mx+b,$ then it is not a linear function.
The given equation has the following characteristics:
\begin{array}{l}\require{cancel}
y \text{ is NOT a function of }x
\\\text{Domain: }
\left[ 0,\infty \right)
\\\text{NOT a linear function}
.\end{array}