1. **Determine domain, range, and function status for each relation:** **a)** Relation: $\{(-3,0), (-1,1), (0,1), (4,5), (0,6)\}$ - Domain: The set of all first elements: $\{-3, -1, 0, 4\}$ - 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 $(-\infty, \infty)$ - Range: All real numbers $(-\infty, \infty)$ - Function? Yes, because for each $x$ there is exactly one $y$. **c)** Graph: V-shaped connected line with points $(-4,0)$ to $(0,-4)$ to $(4,4)$ - Domain: $[-4,4]$ - Range: $[-4,4]$ - Function? Yes, because for each $x$ in the domain, there is exactly one $y$ value. **d)** Relation: $x^2 + y^2 = 16$ - Domain: $[-4,4]$ (since $x^2 \leq 16$) - Range: $[-4,4]$ - Function? No, because for some $x$ values (e.g., $x=0$), there are two $y$ values (positive and negative). 2. **Rule to determine if a relation is a function from its graph:** - Use the **vertical line test**: If any vertical line intersects the graph more than once, the relation is not a function. **Check each relation:** **a)** $\{(-2,1), (1,1), (0,0), (1,-1), (1,-2), (2,-2)\}$ - Domain: $\{-2, 0, 1, 2\}$ - Function? No, because $x=1$ has multiple $y$ values. **b)** $y = 4 - 3x$ - Function? Yes, linear function. **c)** $y = (x-2)^2 + 4$ - Function? Yes, parabola opening upward. **d)** $x^2 + y^2 = 1$ - Function? No, circle fails vertical line test. **e)** $y = \frac{1}{x}$ - Function? Yes, hyperbola with domain $x \neq 0$. **f)** $y = \sqrt{x}$ - Function? Yes, domain $x \geq 0$. 6. Given $f(x) = x^2 - 4x + 3$, find $x$ such that $f(x) = 8$. 1. Set equation: $x^2 - 4x + 3 = 8$ 2. Simplify: $x^2 - 4x + 3 - 8 = 0 \Rightarrow x^2 - 4x - 5 = 0$ 3. Factor or use quadratic formula: $$x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2} = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2}$$ 4. Solutions: $$x = \frac{4 + 6}{2} = 5, \quad x = \frac{4 - 6}{2} = -1$$ 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 descending 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: - Multiply first by 3: $12a + 6b = 60$ - Subtract second: $(12a + 6b) - (36a + 6b) = 60 - (-60) \Rightarrow -24a = 120 \Rightarrow a = -5$ - Substitute $a$ 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: $(-\infty, \infty)$ - Range: Since $(x-1)^2 \geq 0$, minimum value 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 fencing problem: Not enough info to solve here, no equation or figure provided.