1. For each relation, determine the domain, range, and whether it is a function. **a)** Relation: $\{(-3,0), (-1,1), (0,1), (4,5), (0,6)\}$ - Domain: The set of all first elements: $\{-3, -1, 0, 4\}$ (note $0$ repeats but counted once) - Range: The set of all second elements: $\{0, 1, 5, 6\}$ - Function? No, because the input $0$ corresponds to two different outputs $1$ and $6$. **b)** Relation: $y = 4 - x$ - Domain: All real numbers $\mathbb{R}$ (no restrictions) - Range: All real numbers $\mathbb{R}$ (since $y$ can take any value depending on $x$) - Function? Yes, each $x$ has exactly one $y$. **c)** Graph: Two V-shaped lines meeting at origin forming an "X" shape. - Domain: $[-4,4]$ (from graph) - Range: $[-4,4]$ (from graph) - Function? No, fails vertical line test because for some $x$ values (e.g., $x=0$), there are two $y$ values. **d)** Relation: $x^2 + y^2 = 16$ - Domain: $[-4,4]$ (circle radius 4) - Range: $[-4,4]$ - Function? No, circle fails vertical line test (each $x$ has two $y$ values except at edges). 2. Rule to determine if a relation is a function from its graph: The **vertical line test**. If any vertical line intersects the graph more than once, the relation is not a function. Check each: **a)** $\{(-2,1), (1,1), (0,0), (1,-1), (1,-2), (2,-2)\}$ - Domain: $\{-2,0,1,2\}$ - Multiple outputs for $x=1$ ($1, -1, -2$) - Function? No. **b)** $y = 4 - 3x$ - Linear function, passes vertical line test - Function? Yes. **c)** $y = (x-2)^2 + 4$ - Parabola opening upward - Function? Yes. **d)** $x^2 + y^2 = 1$ - Circle of radius 1 - Function? No. **e)** $y = \frac{1}{x}$ - Hyperbola, passes vertical line test - Function? Yes. **f)** $y = \sqrt{x}$ - Domain: $[0, \infty)$ - Function? Yes. 6. Given $f(x) = x^2 - 4x + 3$, find $x$ such that $f(x) = 8$. - Set equation: $x^2 - 4x + 3 = 8$ - Simplify: $x^2 - 4x + 3 - 8 = 0 \Rightarrow x^2 - 4x - 5 = 0$ - Factor or use quadratic formula: $$x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2} = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2}$$ - Solutions: - $x = \frac{4 + 6}{2} = 5$ - $x = \frac{4 - 6}{2} = -1$ 7. Ball thrown upward from 60 m roof, reaches 80 m at 2 s, hits ground at 6 s. **a)** Sketch graph: Height vs time is a parabola starting at 60 m at $t=0$, peaking above 80 m, then falling to 0 at $t=6$. **b)** Domain: $[0,6]$ seconds (time from throw to hitting ground) Range: $[0, H_{max}]$ meters, where $H_{max} \geq 80$ **c)** Equation form: $h(t) = at^2 + bt + c$ - At $t=0$, $h=60 \Rightarrow c=60$ - At $t=2$, $h=80 \Rightarrow 4a + 2b + 60 = 80 \Rightarrow 4a + 2b = 20$ - At $t=6$, $h=0 \Rightarrow 36a + 6b + 60 = 0 \Rightarrow 36a + 6b = -60$ Solve system: - From $4a + 2b = 20$, multiply by 3: $12a + 6b = 60$ - Subtract from $36a + 6b = -60$: $(36a - 12a) + (6b - 6b) = -60 - 60 \Rightarrow 24a = -120 \Rightarrow a = -5$ - Substitute $a=-5$ into $4a + 2b = 20$: $4(-5) + 2b = 20 \Rightarrow -20 + 2b = 20 \Rightarrow 2b = 40 \Rightarrow b = 20$ Equation: $h(t) = -5t^2 + 20t + 60$ 8. Domain and range: **a)** $f(x) = 2(x-1)^2 + 3$ - Domain: $\mathbb{R}$ - Range: Since parabola opens up, minimum at $x=1$ is $3$, so range $[3, \infty)$ **b)** $f(x) = \sqrt{2x + 4}$ - Domain: $2x + 4 \geq 0 \Rightarrow x \geq -2$ - Range: $[0, \infty)$ 9. Farmer has 540 m fencing to enclose rectangle divided into two equal sections by a vertical fence. - Let length = $L$, width = $W$ - Total fencing: $2L + 3W = 540$ (two widths plus three lengths including dividing fence)