which piecewise relation defines a function?\ny=\begin{cases}x^{2},&x < - 2\\0,&-2leq xleq4\\-x^{2},&xgeq4end…

Explanation:

Step1: Recall function - definition

A relation is a function if for each input $x$ - value there is exactly one output $y$ - value.

Step2: Check the first option

The first option has a contradiction for $x = 4$ since $y = 0$ when $-2\leq x\leq4$ and $y=-x^{2}=-16$ when $x\geq4$. So, it's not a function.

Step3: Check the second option

The second option has a contradiction for $x = 2$ since $y = 4$ when $-2\lt x\leq2$ and $y=x^{2}+1 = 5$ when $x\geq2$. So, it's not a function.

Step4: Check the third option

For $x\lt - 2$, $y=-3x$. For $0\leq x\lt4$, $y = 3$. For $x\geq4$, $y = 2x$. There is no overlap in the domains that would cause a single $x$ - value to have more than one $y$ - value. So, it is a function.

Step5: Check the fourth option

The intervals $x\leq - 4$ and $-5\lt x\lt1$ overlap at $-5\lt x\leq - 4$. For example, if $x=-4$, $y=-3x = 12$ from the first - part and $y = 3$ from the second - part. So, it's not a function.

Answer:

$y=\begin{cases}-3x,x\lt - 2\3,0\leq x\lt4\2x,x\geq4\end{cases}$