## Intermediate Algebra (12th Edition)

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