which function is undefined for x = 0?\n$y=sqrt3{x - 2}$\n$y=sqrt{x - 2}$\n$y=sqrt3{x + 2}$\n$y=sqrt{x + 2}$

Explanation:

Step1: Recall square - root and cube - root properties

The cube - root function $\sqrt[3]{a}$ is defined for all real values of $a$, while the square - root function $\sqrt{a}$ is defined only for $a\geq0$.

Step2: Substitute $x = 0$ into each function

For $y=\sqrt[3]{x - 2}$, when $x = 0$, $y=\sqrt[3]{0 - 2}=\sqrt[3]{-2}$, which is defined. For $y=\sqrt{x - 2}$, when $x = 0$, $y=\sqrt{0 - 2}=\sqrt{-2}$, which is undefined in the set of real numbers since the number inside the square - root is negative. For $y=\sqrt[3]{x + 2}$, when $x = 0$, $y=\sqrt[3]{0+2}=\sqrt[3]{2}$, which is defined. For $y=\sqrt{x + 2}$, when $x = 0$, $y=\sqrt{0 + 2}=\sqrt{2}$, which is defined.

Answer:

$y=\sqrt{x - 2}$