which graph represents $y = sqrt3{x}$?

Explanation:

Step1: Analyze the domain

The cube - root function $y = \sqrt[3]{x}$ has a domain of all real numbers, i.e., $x\in(-\infty,\infty)$.

Step2: Analyze key points

When $x = 0$, $y=\sqrt[3]{0}=0$; when $x = 1$, $y=\sqrt[3]{1}=1$; when $x=-1$, $y=\sqrt[3]{-1}=-1$. The graph of $y = \sqrt[3]{x}$ is symmetric about the origin.

Step3: Compare with the given graph

The given graph is a parabola - like shape opening upwards with vertex at the origin, which is the graph of $y = x^{2}$ (not $y=\sqrt[3]{x}$). The graph of $y=\sqrt[3]{x}$ is a smooth curve passing through the origin, increasing for all $x\in(-\infty,\infty)$. The correct graph of $y = \sqrt[3]{x}$ has a different shape from the one shown.

The graph shown does not represent $y=\sqrt[3]{x}$. The graph of $y=\sqrt[3]{x}$ is a smooth curve passing through the origin, symmetric about the origin, and increasing for all real - valued $x$.