1. **Problem 1.1**: Given the quadratic function $$f(x) = -x^2 - 2x - 1$$ 1. Find the x- and y-intercepts. 2. Find the turning point. 3. Sketch the graph. 4. Determine domain and range. **Step 1: Find intercepts** - **y-intercept**: Set $$x=0$$, then $$f(0) = -0 - 0 - 1 = -1$$, so y-intercept is $$(0, -1)$$. - **x-intercepts**: Set $$f(x) = 0$$: $$-x^2 - 2x - 1 = 0 \implies x^2 + 2x + 1 = 0$$ This factors as $$(x+1)^2 = 0$$, so $$x = -1$$ (double root). Thus, x-intercept is $$(-1, 0)$$. **Step 2: Find turning point** - The vertex of $$f(x) = ax^2 + bx + c$$ is at $$x = -\frac{b}{2a}$$. - Here, $$a = -1$$, $$b = -2$$, so $$x = -\frac{-2}{2(-1)} = \frac{2}{-2} = -1$$. - Find $$f(-1)$$: $$f(-1) = -(-1)^2 - 2(-1) - 1 = -1 + 2 - 1 = 0$$. - Turning point is $$(-1, 0)$$. **Step 3: Sketch graph** - Parabola opens downward (since $$a = -1 < 0$$). - Vertex at $$(-1, 0)$$. - y-intercept at $$(0, -1)$$. - x-intercept at $$(-1, 0)$$ (vertex). **Step 4: Domain and range** - Domain of any quadratic: $$(-\infty, \infty)$$. - Range: Since parabola opens downward and vertex is maximum at $$y=0$$, range is $$(-\infty, 0]$$. --- 2. **Problem 1.2**: Given - Total cost: $$TC = 200 + 3Q$$ - Demand (price): $$P = 107 - 2Q$$ **Step 1: Total revenue (TR)** - $$TR = P \times Q = (107 - 2Q)Q = 107Q - 2Q^2$$. **Step 2: Graph TC and TR and find break-even points** - Break-even points occur where $$TC = TR$$: $$200 + 3Q = 107Q - 2Q^2$$ Rearranged: $$0 = 107Q - 2Q^2 - 3Q - 200 = -2Q^2 + 104Q - 200$$ Multiply both sides by $$-1$$: $$2Q^2 - 104Q + 200 = 0$$ - Divide by 2: $$Q^2 - 52Q + 100 = 0$$ - Use quadratic formula: $$Q = \frac{52 \pm \sqrt{52^2 - 4 \times 1 \times 100}}{2} = \frac{52 \pm \sqrt{2704 - 400}}{2} = \frac{52 \pm \sqrt{2304}}{2}$$ $$\sqrt{2304} = 48$$ - So, $$Q = \frac{52 \pm 48}{2}$$ - Two solutions: $$Q_1 = \frac{52 + 48}{2} = \frac{100}{2} = 50$$ $$Q_2 = \frac{52 - 48}{2} = \frac{4}{2} = 2$$ **Step 3: Find corresponding prices** - For $$Q=2$$: $$P = 107 - 2(2) = 107 - 4 = 103$$ - For $$Q=50$$: $$P = 107 - 2(50) = 107 - 100 = 7$$ **Step 4: Interpretation** - Break-even points at $$(2, 103)$$ and $$(50, 7)$$. --- 3. **Problem 2.1**: Simplify $$\frac{8^x \times 6^{x-3} \times 9^{1-x}}{16^{x-1} \times 3^{-x}}$$ **Step 1: Express bases as powers of primes** - $$8 = 2^3$$ - $$6 = 2 \times 3$$ - $$9 = 3^2$$ - $$16 = 2^4$$ Rewrite: $$\frac{(2^3)^x \times (2 \times 3)^{x-3} \times (3^2)^{1-x}}{(2^4)^{x-1} \times 3^{-x}} = \frac{2^{3x} \times 2^{x-3} 3^{x-3} \times 3^{2(1-x)}}{2^{4(x-1)} \times 3^{-x}}$$ **Step 2: Combine powers of 2 and 3** - Powers of 2 numerator: $$2^{3x} \times 2^{x-3} = 2^{3x + x - 3} = 2^{4x - 3}$$ - Powers of 3 numerator: $$3^{x-3} \times 3^{2 - 2x} = 3^{x - 3 + 2 - 2x} = 3^{-x - 1}$$ - Denominator: $$2^{4(x-1)} \times 3^{-x} = 2^{4x - 4} \times 3^{-x}$$ **Step 3: Divide powers** - For 2: $$2^{4x - 3} / 2^{4x - 4} = 2^{(4x - 3) - (4x - 4)} = 2^{1} = 2$$ - For 3: $$3^{-x - 1} / 3^{-x} = 3^{(-x - 1) - (-x)} = 3^{-1} = \frac{1}{3}$$ **Step 4: Final simplified expression** $$2 \times \frac{1}{3} = \frac{2}{3}$$ --- 4. **Problem 2.2a**: Solve $$3 \times 7^{x+3} = 12$$ **Step 1: Isolate exponential term** $$7^{x+3} = \frac{12}{3} = 4$$ **Step 2: Take logarithm base 7** $$x + 3 = \log_7 4$$ **Step 3: Solve for x** $$x = \log_7 4 - 3$$ --- 5. **Problem 2.2b**: Solve $$\log_3(x - 1) + \log_3(x + 1) = 2$$ **Step 1: Use log property** $$\log_3[(x - 1)(x + 1)] = 2$$ **Step 2: Simplify inside log** $$(x - 1)(x + 1) = x^2 - 1$$ **Step 3: Convert log equation to exponential** $$x^2 - 1 = 3^2 = 9$$ **Step 4: Solve quadratic** $$x^2 - 1 = 9 \implies x^2 = 10 \implies x = \pm \sqrt{10}$$ **Step 5: Check domain** - Arguments of logs must be positive: $$x - 1 > 0 \implies x > 1$$ $$x + 1 > 0 \implies x > -1$$ - So only $$x = \sqrt{10}$$ is valid. --- 6. **Problem 2.3a**: Population after 15 years $$P = 350 e^{0.05t}$$ Calculate for $$t=15$$: $$P = 350 e^{0.05 \times 15} = 350 e^{0.75}$$ Using $$e^{0.75} \approx 2.117$$: $$P \approx 350 \times 2.117 = 740.95$$ --- 7. **Problem 2.3b**: Year population reaches 950 Set $$P = 950$$: $$950 = 350 e^{0.05t}$$ Divide both sides: $$\frac{950}{350} = e^{0.05t} \implies \frac{19}{7} = e^{0.05t}$$ Take natural log: $$\ln \frac{19}{7} = 0.05t$$ Calculate $$\ln \frac{19}{7} \approx \ln 2.714 = 0.999$$ Solve for $$t$$: $$t = \frac{0.999}{0.05} = 19.98 \approx 20$$ years after 1995 Year = 1995 + 20 = 2015 --- 8. **Problem 2.4**: Find limit $$\lim_{x \to 6} \frac{2x - 12}{x^2 + x - 42}$$ **Step 1: Simplify numerator and denominator** - Numerator: $$2x - 12 = 2(x - 6)$$ - Denominator: $$x^2 + x - 42 = (x - 6)(x + 7)$$ **Step 2: Cancel common factor** $$\frac{2(x - 6)}{(x - 6)(x + 7)} = \frac{2}{x + 7}, \quad x \neq 6$$ **Step 3: Evaluate limit** $$\lim_{x \to 6} \frac{2}{x + 7} = \frac{2}{6 + 7} = \frac{2}{13}$$ --- Final answers: - 1.1 a) y-intercept: (0, -1), x-intercept: (-1, 0) - 1.1 b) Turning point: (-1, 0) - 1.1 d) Domain: $$(-\infty, \infty)$$, Range: $$(-\infty, 0]$$ - 1.2 a) $$TR = 107Q - 2Q^2$$ - 1.2 b) Break-even points: $$(2, 103)$$ and $$(50, 7)$$ - 2.1 Simplified expression: $$\frac{2}{3}$$ - 2.2 a) $$x = \log_7 4 - 3$$ - 2.2 b) $$x = \sqrt{10}$$ - 2.3 a) Population after 15 years: 740.95 - 2.3 b) Year population reaches 950: 2015 - 2.4 Limit: $$\frac{2}{13}$$