1. **Problem Statement:** We are given two relations: a. $y = -2$ b. $y = \pm\sqrt{x} + 1$ We need to create tables of values, plot the relations, and determine if each is a function using the Vertical Line Test (VLT). --- 2. **Relation a: $y = -2$** - This is a horizontal line where $y$ is always $-2$ regardless of $x$. - Table of values (choose $x = -2, -1, 0, 1, 2$): $$\begin{array}{c|c} x & y \\\hline -2 & -2 \\ -1 & -2 \\ 0 & -2 \\ 1 & -2 \\ 2 & -2 \end{array}$$ - Since for every $x$ there is exactly one $y$, it passes the VLT and is a function. --- 3. **Relation b: $y = \pm\sqrt{x} + 1$** - This means $y = +\sqrt{x} + 1$ and $y = -\sqrt{x} + 1$. - Domain: $x \geq 0$ because square root is defined for non-negative $x$. - Table of values for $x = 0, 1, 4, 9, 16$: $$\begin{array}{c|cc} x & y = +\sqrt{x} + 1 & y = -\sqrt{x} + 1 \\\hline 0 & 1 & 1 \\ 1 & 2 & 0 \\ 4 & 3 & -1 \\ 9 & 4 & -2 \\ 16 & 5 & -3 \end{array}$$ - Since for some $x$ values (e.g., $x=1$) there are two different $y$ values, it fails the VLT and is not a function. --- 4. **Summary:** - Relation a is a function and its graph is a horizontal line at $y = -2$. - Relation b is not a function because it fails the VLT due to two $y$ values for some $x$. --- **Final answers:** - a) Function, horizontal line $y = -2$. - b) Not a function, graph is two branches $y = +\sqrt{x} + 1$ and $y = -\sqrt{x} + 1$.