1. Determine the domain, range, and whether the relation is a function for each given relation. 1.a) Relation: ${\{(-3,0), (-1,1), (0,1), (4,5), (0,6)\}}$ - Domain: all x-values ${\{-3, -1, 0, 4\}}$ - Range: all y-values ${\{0, 1, 5, 6\}}$ - Is it a function? No, because the input $0$ maps to two different outputs ($1$ and $6$). 1.b) Relation: $y = 4 - x$ - Domain: All real numbers (since $x$ can be anything) - Range: All real numbers (linear function) - Is it a function? Yes, each $x$ has exactly one $y$. 1.c) Relation from table: \begin{tabular}{c|ccccc} x & -4 & -2 & 0 & 2 & 4 \\ \hline y & 4 & 2 & 0 & -2 & -4 \end{tabular} - Domain: ${\{-4, -2, 0, 2, 4\}}$ - Range: ${\{4, 2, 0, -2, -4\}}$ - Is it a function? Yes, each input has one output. 1.d) Relation: $x^2 + y^2 = 16$ - Domain: ${x \in [-4,4]}$ (since $|x| \le 4$ for real $y$) - Range: ${y \in [-4,4]}$ - Is it a function? No, because for some $x$ there are two $y$ values (circle). 2. Rule to determine function from graph: A relation is a function if any vertical line intersects the graph at most once (Vertical Line Test). 2.a) Relation: ${(-2,1), (1,1), (0,0), (1,-1), (1,-2), (2,-2)}$ - x=1 maps to three different y values, so not a function. 2.b) $y = 4 - 3x$ - Linear function, passes vertical line test, so it's a function. 2.c) $y = (x - 2)^2 + 4$ - Parabola opening upward, passes vertical line test, so it is a function. 2.d) $x^2 + y^2 = 1$ - Circle with radius 1, fails vertical line test, not a function. 2.e) $y = \frac{1}{x}$ - Hyperbola, passes vertical line test except at $x=0$ (not in domain), so it's a function. 2.f) $y = \sqrt{x}$ - Passes vertical line test, domain $x\ge0$, so it's a function. Summary: - Functions: 1.b, 1.c, 2.b, 2.c, 2.e, 2.f - Not functions: 1.a, 1.d, 2.a, 2.d