Let's solve each question step-by-step: --- ### **1. Simplify \(\frac{x^2-4}{x^2-9} - \frac{x-3}{x+2}\)** **Step 1:** Factor numerators and denominators where possible: - \(x^2-4 = (x-2)(x+2)\) - \(x^2-9 = (x-3)(x+3)\) So, \[ \frac{x^2-4}{x^2-9} = \frac{(x-2)(x+2)}{(x-3)(x+3)} \] **Step 2:** Find a common denominator for both terms. The denominator for the second term is \(x+2\). So, write it as: \[ \frac{x-3}{x+2} = \frac{(x-3)(x+3)}{(x+2)(x+3)} \] But we want a common denominator of \((x-3)(x+3)\), so: \[ \frac{x-3}{x+2} = \frac{(x-3)(x+3)}{(x+2)(x+3)} \] But this doesn't match the denominator we want. Let's try multiplying numerator and denominator by \((x-3)(x+3)\): \[ \frac{x-3}{x+2} = \frac{(x-3)(x+3)}{(x+2)(x+3)} \] But the denominators are \((x-3)(x+3)\) and \((x+2)\). We need a common denominator: \((x-3)(x+3)\). \[ \frac{x-3}{x+2} = \frac{(x-3)(x+3)}{(x+2)(x+3)} \] But now, let's express both terms with denominator \((x-3)(x+3)\): So, for \(\frac{x-3}{x+2}\), multiply numerator and denominator by \((x+2)\): \[ \frac{x-3}{x+2} = \frac{(x-3)(x+3)}{(x+2)(x+3)} \] Wait, let's do it directly: \[ \frac{x^2-4}{x^2-9} - \frac{x-3}{x+2} \] \[ = \frac{(x-2)(x+2)}{(x-3)(x+3)} - \frac{x-3}{x+2} \] Find common denominator: \((x-3)(x+3)\). Express \(\frac{x-3}{x+2}\) with denominator \((x-3)(x+3)\): \[ \frac{x-3}{x+2} = \frac{(x-3)(x+3)}{(x+2)(x+3)} \] But this doesn't work. Let's multiply numerator and denominator by \((x+3)\): \[ \frac{x-3}{x+2} = \frac{(x-3)(x+3)}{(x+2)(x+3)} \] Now, both have denominator \((x+2)(x+3)\). But the first term has denominator \((x-3)(x+3)\). Find a common denominator: \((x-3)(x+3)(x+2)\). Rewrite both terms with this denominator: \[ \frac{(x-2)(x+2)}{(x-3)(x+3)} = \frac{(x-2)(x+2)(x+2)}{(x-3)(x+3)(x+2)} \] \[ \frac{x-3}{x+2} = \frac{(x-3)(x+3)}{(x+2)(x+3)} = \frac{(x-3)(x+3)(x-3)}{(x-3)(x+3)(x+2)} \] Now, combine: \[ \frac{(x-2)(x+2) - (x-3)(x+3)}{(x-3)(x+3)} \] Expanding numerator: \[ (x-2)(x+2) = x^2 - 4 \] \[ (x-3)(x+3) = x^2 - 9 \] So numerator: \[ (x^2 - 4) - (x^2 - 9) = x^2 - 4 - x^2 + 9 = 5 \] So the answer is \(\frac{5}{x^2-9}\) or \(\frac{5}{(x-3)(x+3)}\). This matches **option D**: \(\frac{5}{x^2-9}\). --- ### **2. Simplify \((x^2-3x+2)(3x+1)\)** First, factor \(x^2-3x+2 = (x-1)(x-2)\), but let's expand directly: \[ (x^2-3x+2)(3x+1) = x^2(3x+1) - 3x(3x+1) + 2(3x+1) \] \[ = 3x^3 + x^2 - 9x^2 - 3x + 6x + 2 \] \[ = 3x^3 + x^2 - 9x^2 - 3x + 6x + 2 \] Combine like terms: \[ 3x^3 + (x^2 - 9x^2) + (-3x + 6x) + 2 \] \[ = 3x^3 - 8x^2 + 3x + 2 \] This matches **option C**. --- ### **3. Simplify \(\frac{4a^2b^2}{a^{3/2}b^{1/2}}\)** Apply the laws of exponents: \[ = 4 \cdot a^{2 - 3/2} \cdot b^{2 - 1/2} \] \[ = 4 \cdot a^{4/2 - 3/2} \cdot b^{4/2 - 1/2} \] \[ = 4 \cdot a^{1/2} \cdot b^{3/2} \] So, \(4a^{1/2}b^{3/2}\) This matches **option D**. --- ### **4. If \(1 \leq x \leq 3\), what is \(|x-1| + |x-2|\)?** - For \(x=1\): \(|1-1| + |1-2| = 0 + 1 = 1\) - For \(x=2\): \(|2-1| + |2-2| = 1 + 0 = 1\) - For \(x=3\): \(|3-1| + |3-2| = 2 + 1 = 3\) But the minimum value is 1, maximum is 3. However, the options are constants, so let's check the sum for all integer values: The only value that fits is **option B: \(x+4\)**, but let's check this for \(x=1\): \(1+4=5\); that's not correct. Wait, let's check the option that fits all values. From above, the sum is always 1 for \(x=1\) and \(x=2\), and 3 for \(x=3\). But none of the options match. The only negative option is C: -7, which is not possible. Let's check option B: \(x+4\). For \(x=1\), \(1+4=5\); doesn't match. For \(x=2\), \(2+4=6\); doesn't match. For \(x=3\), \(3+4=7\); doesn't match. Try option D: \(+3\). For \(x=3\), equals 3; matches for \(x=3\). Option A: \(-3\) is not possible. Thus, for \(x=1\) and \(x=2\) the sum is 1, for \(x=3\) it is 3. None of the options fit correctly. But the only possible answer is **Option D: +3** (for \(x=3\); perhaps a typo in the question). --- ### **5. Set Operations** Given: - \(A = \{1,2,5,6,7,3\}\) - \(B = \{-3,1,2,4,5,7,11\}\) - \(C = \{2,3,7,11,13\}\) - \(D = \{2,5,7,8,9,11,12\}\) Find \((A \cap B) \cup (C \cap D)\): **Step 1: \(A \cap B\):** Elements common to both: A: 1,2,5,6,7,3 B: -3,1,2,4,5,7,11 Common: 1,2,5,7 So, \(A \cap B = \{1,2,5,7\}\) **Step 2: \(C \cap D\):** Elements common to both: C: 2,3,7,11,13 D: 2,5,7,8,9,11,12 Common: 2,7,11 So, \(C \cap D = \{2,7,11\}\) **Step 3: Union** \((A \cap B) \cup (C \cap D)\) Combine: 1,2,5,7,11 So, \(\{1,2,5,7,11\}\) This matches **option B**. --- ### **Final Answers Table** | Question Number | 1 | 2 | 3 | 4 | 5 | |-----------------|---|---|---|---|---| | Answer (letter) | D | C | D | D | B | --- **Final Answers:** 1. D 2. C 3. D 4. D 5. B

Let's carefully analyze and solve each question based on the corrected information. --- ## **3. Simplify \(\frac{4a^3b^2}{a^2b^3}\)** ### Step 1: Apply the laws of exponents: \[ \frac{4a^3b^2}{a^2b^3} = 4 \times a^{3-2} \times b^{2-3} \] ### Step 2: Simplify the exponents: \[ = 4 \times a^{1} \times b^{-1} \] ### Step 3: Write with positive exponents: \[ = 4a \times \frac{1}{b} = \frac{4a}{b} \] ### **Answer: Option B: \(\frac{4a}{b}\)** --- ## **4. For \(1 \leq x \leq 3\), find the value of \(|x - 1| + |x - 2|\)** Let's analyze the behavior of the expression over the interval: - For \(x\) in \([1,2]\): - \(|x - 1|\) = \(x - 1\) - \(|x - 2|\) = \(2 - x\) Sum: \[ (x - 1) + (2 - x) = 1 \] - For \(x\) in \([2,3]\): - \(|x - 1|\) = \(x - 1\) - \(|x - 2|\) = \(x - 2\) Sum: \[ (x - 1) + (x - 2) = 2x - 3 \] At \(x=2\): \[ 2(2) - 3 = 4 - 3 = 1 \] At \(x=3\): \[ 2(3) - 3 = 6 - 3 = 3 \] **Summary:** - For \(x \in [1,2]\), the sum is **constant 1**. - For \(x \in [2,3]\), the sum increases from 1 to 3. ### **Thus, the value of \(|x - 1| + |x - 2|\) over the interval is:** - Constant at 1 between 1 and 2. - Increases to 3 at 3. **Matching options:** - Option A: \(2 - 3\) (which is \(-1\)), incorrect. - Option B: \(x + 4\), ranges from 5 to 7, not matching. - Option C: \(-7\), no. - **Option D: \(+3\)**, matches the maximum at \(x=3\), but the sum is 1 over \([1,2]\). Since the question asks generally for the value in the interval, the most consistent answer, considering the sum is **constant 1** over \([1,2]\), and then increases, the best match is: **Answer: D. +3** (since at \(x=3\), the sum is 3). --- ## **5. Set operation: \((A \cap B) \cup (C \cap D)\)** Given: - \(A = \{1, 2, 5, 6, 7, 3\}\) - \(B = \{-3, 1, 2, 4, 5, 7, 11\}\) - \(C = \{2, 3, 7, 11, 13\}\) - \(D = \{2, 5, 7, 8, 9, 11, 12\}\) ### Step 1: Find \(A \cap B\): - Common elements: \[ \{1, 2, 5, 7\} \] ### Step 2: Find \(C \cap D\): - Common elements: \[ \{2, 7, 11\} \] ### Step 3: Find the union: \[ \{1, 2, 5, 7\} \cup \{2, 7, 11\} = \{1, 2, 5, 7, 11\} \] ### **Answer: Option B: \(\{2, 5, 7, 11\}\)** --- # **Summary of final answers:** | Question | Answer | |------------|---------| | 3 | B. \(\frac{4a}{b}\) | | 4 | D. \(+3\) | | 5 | B. \(\{2, 5, 7, 11\}\) | --- **Note:** The reasoning aligns with the corrected problem statements.

Let's interpret and solve the problem step-by-step. --- ## **Express each radical in simplest form and perform the indicated operations** Given: \[ \sqrt{7}, \quad 7^{15/14} \] --- ### **Step 1: Simplify \(\sqrt{7}\)** \[ \sqrt{7} = 7^{1/2} \] --- ### **Step 2: Simplify \(7^{15/14}\)** This is already in exponential form; no further simplification needed. --- ### **Step 3: Write the expression** \[ \sqrt{7} - 7^{15/14} = 7^{1/2} - 7^{15/14} \] --- ### **Step 4: Find a common base and compare exponents** Express both as powers with base 7: - \(7^{1/2}\) - \(7^{15/14}\) Note that: \[ \frac{1}{2} = \frac{7}{14} \] because: \[ \frac{1}{2} = \frac{7}{14} \] --- ### **Step 5: Rewrite \(7^{1/2}\) as \(7^{7/14}\)** \[ 7^{1/2} = 7^{7/14} \] Now, the expression becomes: \[ 7^{7/14} - 7^{15/14} \] --- ### **Step 6: Factor out the common base** Since both are powers of 7 with exponents \(7/14\) and \(15/14\), factor out \(7^{7/14}\): \[ 7^{7/14} \left(1 - 7^{(15/14) - (7/14)}\right) \] Calculate the exponent in the parentheses: \[ (15/14) - (7/14) = 8/14 = 4/7 \] Thus: \[ 7^{7/14} \left(1 - 7^{4/7}\right) \] --- ### **Final simplified form:** \[ \boxed{ 7^{7/14} \left(1 - 7^{4/7}\right) } \] --- ## **Answer:** \[ \boxed{ \sqrt{7} - 7^{15/14} = 7^{7/14} \left(1 - 7^{4/7}\right) } \]

Let's clarify and solve the given expression step-by-step: --- ## **Expression:** \[ \sqrt[7]{\sqrt{7}} - \sqrt[14]{7^{15}} \] --- ## **Step 1: Simplify \(\sqrt[7]{\sqrt{7}}\)** - \(\sqrt{7} = 7^{1/2}\) - \(\sqrt[7]{\sqrt{7}} = (7^{1/2})^{1/7}\) Using the power of a power rule: \[ = 7^{(1/2) \times (1/7)} = 7^{1/14} \] --- ## **Step 2: Simplify \(\sqrt[14]{7^{15}}\)** - \(\sqrt[14]{7^{15}} = (7^{15})^{1/14}\) Applying the power rule: \[ = 7^{15/14} \] --- ## **Step 3: Write the simplified expression** \[ 7^{1/14} - 7^{15/14} \] --- ## **Step 4: Factor out the common base** Notice that: \[ 7^{1/14} = 7^{1/14} \] \[ 7^{15/14} = 7^{1/14 + 14/14} = 7^{1/14} \times 7^{14/14} = 7^{1/14} \times 7^{1} \] So, rewrite the expression: \[ 7^{1/14} - 7^{1/14} \times 7^{1} \] Factor out \(7^{1/14}\): \[ 7^{1/14} \left(1 - 7^{1}\right) \] Since \(7^{1} = 7\): \[ 7^{1/14} (1 - 7) = 7^{1/14} \times (-6) \] --- ## **Final answer:** \[ \boxed{ -6 \times 7^{1/14} } \] This is the simplest form of the given expression.

To solve this problem, we will use the formulas for calculating the monthly payment and the remaining balance of a mortgage. ### Given Data - **Principal (P)**: $200,000 - **Annual Interest Rate (r)**: 3% (0.03) - **Monthly Interest Rate (i)**: \( \frac{0.03}{12} = 0.0025 \) - **Total Number of Payments (n)**: 30 years = 360 months - **Payments for the first 5 years**: 5 years = 60 months ### Step 1: Calculate the Monthly Payment The formula for the monthly payment (M) for a loan is given by: \[ M = P \cdot \frac{i(1+i)^n}{(1+i)^n - 1} \] Where: - \(P\) is the loan amount, - \(i\) is the monthly interest rate, - \(n\) is the total number of payments. Substituting the values: \[ M = 200,000 \cdot \frac{0.0025(1+0.0025)^{360}}{(1+0.0025)^{360} - 1} \] Calculate \( (1 + 0.0025)^{360} \): \[ (1.0025)^{360} \approx 2.89828 \] Now, substitute this back into the monthly payment formula: \[ M = 200,000 \cdot \frac{0.0025 \cdot 2.89828}{2.89828 - 1} \] \[ M = 200,000 \cdot \frac{0.0072457}{1.89828} \] \[ M \approx 200,000 \cdot 0.0038182 \approx 763.64 \] ### Monthly Payment \[ \text{Monthly Payment} \approx \boxed{763.64} \] ### Step 2: Calculate the Balloon Payment To find the balloon payment at the end of 5 years, we need to calculate the remaining balance after 60 payments. The formula for the remaining balance (B) after \(k\) payments is: \[ B = P \cdot \frac{(1 + i)^n - (1 + i)^k}{(1 + i)^n - 1} \] Where: - \(k\) is the number of payments made (60). Substituting the values: \[ B = 200,000 \cdot \frac{(1 + 0.0025)^{360} - (1 + 0.0025)^{60}}{(1 + 0.0025)^{360} - 1} \] Calculate \( (1.0025)^{60} \): \[ (1.0025)^{60} \approx 1.1616 \] Now substituting back into the formula: \[ B = 200,000 \cdot \frac{2.89828 - 1.1616}{2.89828 - 1} \] \[ B = 200,000 \cdot \frac{1.73668}{1.89828} \approx 200,000 \cdot 0.9156 \approx 183,120 \] ### Balloon Payment \[ \text{Balloon Payment} \approx \boxed{183,120} \] ### Final Summary - **Monthly Payment**: \(\$763.64\) - **Balloon Payment**: \(\$183,120\)

To analyze the polar equation \( r = 2 - 5\cos(\theta) \), we will tackle each part step by step. --- ### **i. Symmetry and Origin** **Symmetry about the z-axis:** A polar graph is symmetric about the z-axis if replacing \(\theta\) with \(-\theta\) does not change the equation. 1. Substitute \(-\theta\) into the equation: \[ r = 2 - 5\cos(-\theta) \] 2. Since \(\cos(-\theta) = \cos(\theta)\): \[ r = 2 - 5\cos(\theta) \] This shows that the graph is symmetric about the z-axis. **Passing through the origin:** To check if the graph passes through the origin, set \(r = 0\): \[ 0 = 2 - 5\cos(\theta) \implies 5\cos(\theta) = 2 \implies \cos(\theta) = \frac{2}{5} \] This equation has valid solutions for \(\theta\) within the range of \(0\) to \(180\) degrees, indicating that the graph indeed passes through the origin. --- ### **ii. Construct a Table for \(r\) and \(\theta\)** We will calculate \(r\) for angles \(\theta\) ranging from \(0^\circ\) to \(180^\circ\). | \(\theta\) (degrees) | \(\theta\) (radians) | \(r = 2 - 5\cos(\theta)\) | |----------------------|----------------------|----------------------------| | 0 | 0 | \(2 - 5 \cdot 1 = -3\) | | 30 | \(\frac{\pi}{6}\) | \(2 - 5 \cdot \frac{\sqrt{3}}{2} \approx -0.33\) | | 60 | \(\frac{\pi}{3}\) | \(2 - 5 \cdot \frac{1}{2} = -0.5\) | | 90 | \(\frac{\pi}{2}\) | \(2 - 5 \cdot 0 = 2\) | | 120 | \(\frac{2\pi}{3}\) | \(2 - 5 \cdot (-\frac{1}{2}) = 4.5\) | | 150 | \(\frac{5\pi}{6}\) | \(2 - 5 \cdot (-\frac{\sqrt{3}}{2}) \approx 5.33\) | | 180 | \(\pi\) | \(2 - 5 \cdot (-1) = 7\) | --- ### **iii. Plotting the Graph** Using a polar grid, plot the points based on the table: - For \(0^\circ\), \(r = -3\) (which means point is in the opposite direction, so at \(180^\circ\) with \(3\) unit). - For \(30^\circ\), \(r \approx -0.33\). - For \(60^\circ\), \(r = -0.5\). - For \(90^\circ\), \(r = 2\). - For \(120^\circ\), \(r = 4.5\). - For \(150^\circ\), \(r \approx 5.33\). - For \(180^\circ\), \(r = 7\). --- ### **iv. Plot \( \theta = 150^\circ \)** 1. For \(\theta = 150^\circ\): - \(r = 2 - 5\cos(150^\circ) = 2 - 5 \cdot (-\frac{\sqrt{3}}{2}) = 5.33\). 2. Find intersection points: - Set the equations equal to find intersections: \( r = 2 - 5\cos(\theta)\) and the line \( \theta = 150^\circ\): \[ r = 5.33 \quad \text{(for } \theta = 150^\circ\text{)} \] ### **Conclusion:** - The graph is symmetric about the z-axis and passes through the origin. - The table lists values necessary to plot the polar graph. - The graph should show a loop because of the negative \(r\) values at angles \(0^\circ\), \(30^\circ\), and \(60^\circ\). - The intersection point at \(\theta = 150^\circ\) indicates that the line intersects the curve, and the coordinates can be plotted accordingly. ### **Final Graph** Use the polar grid to plot all points and the line for \(\theta = 150^\circ\) to show intersections. --- This completes the analysis and steps regarding the polar equation \(r = 2 - 5\cos(\theta)\).

To estimate the Consumer Price Index (CPI) for the years 2014 and 2015 based on the provided data points for the years 2011 and 2016, we can follow these steps: ### **Step 1: Identify the Data Points** - In 2011 (\(x = 11\)), CPI = 229.5 - In 2016 (\(x = 16\)), CPI = 234.2 ### **Step 2: Create the Coordinates** Using the data points: - Point 1: \((11, 229.5)\) - Point 2: \((16, 234.2)\) ### **Step 3: Calculate the Slope (m)** The slope \(m\) of a line through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] Substituting in our values: \[ m = \frac{234.2 - 229.5}{16 - 11} = \frac{4.7}{5} = 0.94 \] ### **Step 4: Write the Equation of the Line** Using the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\), we can use one of the points (11, 229.5): \[ y - 229.5 = 0.94(x - 11) \] Expanding this equation: \[ y - 229.5 = 0.94x - 10.34 \] \[ y = 0.94x + 219.16 \] ### **Final Equation** The linear equation that models the CPI is: \[ y = 0.94x + 219.16 \] ### **Step 5: Estimate CPI for 2014 and 2015** To find the CPI for the years 2014 and 2015, we substitute \(x\) values corresponding to those years. - **For 2014**: \(x = 14\) \[ y = 0.94(14) + 219.16 \] \[ y = 13.16 + 219.16 = 232.32 \] - **For 2015**: \(x = 15\) \[ y = 0.94(15) + 219.16 \] \[ y = 14.1 + 219.16 = 233.26 \] ### **Final Answers** - The CPI in **2014** would be approximately **232.32**. - The CPI in **2015** would be approximately **233.26**. ### **Summary** - **Linear Equation**: \(y = 0.94x + 219.16\) - **CPI in 2014**: \(\boxed{232.32}\) - **CPI in 2015**: \(\boxed{233.26}\)

To analyze the given functions \(f(x)\), \(g(x)\), and \(m(x)\), we will determine their domain, vertical asymptotes, end behavior asymptotes, intercepts, and sketch their graphs. --- ## **1. \(f(x) = \frac{3}{x - 2}\)** ### Domain: - The function is undefined when the denominator is zero. - Set \(x - 2 = 0 \implies x = 2\). - **Domain**: \(x \in (-\infty, 2) \cup (2, \infty)\) ### Vertical Asymptote: - The vertical asymptote occurs where the function is undefined: - **Vertical Asymptote**: \(x = 2\) ### End Behavior Asymptote: - As \(x \to \pm \infty\), \(f(x) \to 0\). - **End Behavior Asymptote**: \(y = 0\) ### Intercepts: - **y-intercept**: Set \(x = 0\): \[ f(0) = \frac{3}{0 - 2} = -1.5 \] - **x-intercept**: No x-intercept since the numerator is a constant. ### Summary: - **Domain**: \((- \infty, 2) \cup (2, \infty)\) - **Vertical Asymptote**: \(x = 2\) - **End Behavior Asymptote**: \(y = 0\) - **y-intercept**: \((0, -1.5)\) ### Graph: The graph approaches the vertical asymptote at \(x=2\) and approaches the horizontal asymptote \(y=0\) as \(x\) approaches \(\pm \infty\). --- ## **2. \(g(x) = \frac{x - 1}{x^2 - 9}\)** ### Domain: - Set the denominator to zero to find undefined points: \[ x^2 - 9 = 0 \implies (x - 3)(x + 3) = 0 \implies x = 3, -3 \] - **Domain**: \(x \in (-\infty, -3) \cup (-3, 3) \cup (3, \infty)\) ### Vertical Asymptotes: - Vertical asymptotes occur at: - **Vertical Asymptotes**: \(x = -3\) and \(x = 3\) ### End Behavior Asymptote: - As \(x \to \pm \infty\), \(g(x) \to 0\). - **End Behavior Asymptote**: \(y = 0\) ### Intercepts: - **y-intercept**: Set \(x = 0\): \[ g(0) = \frac{0 - 1}{0^2 - 9} = \frac{-1}{-9} = \frac{1}{9} \] - **x-intercept**: Set the numerator to zero: \[ x - 1 = 0 \implies x = 1 \] ### Summary: - **Domain**: \((- \infty, -3) \cup (-3, 3) \cup (3, \infty)\) - **Vertical Asymptotes**: \(x = -3, 3\) - **End Behavior Asymptote**: \(y = 0\) - **y-intercept**: \((0, \frac{1}{9})\) - **x-intercept**: \((1, 0)\) ### Graph: The graph approaches the vertical asymptotes at \(x = -3\) and \(x = 3\) and approaches the horizontal asymptote \(y = 0\) as \(x\) approaches \(\pm \infty\). --- ## **3. \(m(x) = \frac{2x}{x + 3}\)** ### Domain: - The function is undefined when \(x + 3 = 0\): \[ x = -3 \] - **Domain**: \(x \in (-\infty, -3) \cup (-3, \infty)\) ### Vertical Asymptote: - The vertical asymptote occurs at: - **Vertical Asymptote**: \(x = -3\) ### End Behavior Asymptote: - As \(x \to \pm \infty\), the function approaches: \[ y = \frac{2x}{x} = 2 \] - **End Behavior Asymptote**: \(y = 2\) ### Intercepts: - **y-intercept**: Set \(x = 0\): \[ m(0) = \frac{2 \cdot 0}{0 + 3} = 0 \] - **x-intercept**: Set the numerator to zero: \[ 2x = 0 \implies x = 0 \] ### Summary: - **Domain**: \((- \infty, -3) \cup (-3, \infty)\) - **Vertical Asymptote**: \(x = -3\) - **End Behavior Asymptote**: \(y = 2\) - **y-intercept**: \((0, 0)\) - **x-intercept**: \((0, 0)\) ### Graph: The graph approaches the vertical asymptote at \(x = -3\) and approaches the horizontal asymptote \(y = 2\) as \(x\) approaches \(\pm \infty\). --- ## **Final Overview of All Functions:** | Function | Domain | Vertical Asymptotes | End Behavior Asymptote | y-intercepts | x-intercepts | |----------|--------|---------------------|-------------------------|--------------|---------------| | \(f(x)\) | \((- \infty, 2) \cup (2, \infty)\) | \(x = 2\) | \(y = 0\) | \((0, -1.5)\) | None | | \(g(x)\) | \((- \infty, -3) \cup (-3, 3) \cup (3, \infty)\) | \(x = -3, 3\) | \(y = 0\) | \((0, \frac{1}{9})\) | \((1, 0)\) | | \(m(x)\) | \((- \infty, -3) \cup (-3, \infty)\) | \(x = -3\) | \(y = 2\) | \((0, 0)\) | \((0, 0)\) | ### Sketching the Graphs: - For \(f(x)\), the graph approaches \(y = 0\) and has a vertical asymptote at \(x = 2\). - For \(g(x)\), the graph has vertical asymptotes at \(x = -3\) and \(x = 3\) and approaches \(y = 0\). - For \(m(x)\), the graph approaches \(y = 2\) with a vertical asymptote at \(x = -3\). This completes the analysis of the functions \(f(x)\), \(g(x)\), and \(m(x)\).

To analyze the function \(t(x) = \frac{2x^2}{x^2 + 2x - 8}\), we will determine its domain, vertical asymptotes, end behavior asymptote, intercepts, and sketch the graph. --- ## **Function Analysis: \(t(x) = \frac{2x^2}{x^2 + 2x - 8}\)** ### **Step 1: Domain** The function is undefined where the denominator equals zero. Thus, we set the denominator to zero and solve for \(x\): \[ x^2 + 2x - 8 = 0 \] Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): - \(a = 1\), \(b = 2\), \(c = -8\) Calculating the discriminant: \[ b^2 - 4ac = 2^2 - 4(1)(-8) = 4 + 32 = 36 \] Now, applying the quadratic formula: \[ x = \frac{-2 \pm \sqrt{36}}{2(1)} = \frac{-2 \pm 6}{2} \] This gives us: \[ x = \frac{4}{2} = 2 \quad \text{and} \quad x = \frac{-8}{2} = -4 \] - **Domain**: \(x \in (-\infty, -4) \cup (-4, 2) \cup (2, \infty)\) ### **Step 2: Vertical Asymptotes** Vertical asymptotes occur at the values of \(x\) that make the denominator zero: - **Vertical Asymptotes**: \(x = -4\) and \(x = 2\) ### **Step 3: End Behavior Asymptote** To find the end behavior asymptote, we analyze the degrees of the numerator and denominator: - The degree of the numerator \(2x^2\) is 2. - The degree of the denominator \(x^2 + 2x - 8\) is also 2. Since the degrees are the same, the end behavior asymptote is given by: \[ y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} = \frac{2}{1} = 2 \] - **End Behavior Asymptote**: \(y = 2\) ### **Step 4: Intercepts** **y-intercept**: Set \(x = 0\): \[ t(0) = \frac{2(0)^2}{(0)^2 + 2(0) - 8} = \frac{0}{-8} = 0 \] Thus, the y-intercept is: - **y-intercept**: \((0, 0)\) **x-intercepts**: Set the numerator to zero: \[ 2x^2 = 0 \implies x = 0 \] Thus, the x-intercept is: - **x-intercept**: \((0, 0)\) ### **Summary of \(t(x)\) Analysis** - **Domain**: \(x \in (-\infty, -4) \cup (-4, 2) \cup (2, \infty)\) - **Vertical Asymptotes**: \(x = -4\) and \(x = 2\) - **End Behavior Asymptote**: \(y = 2\) - **y-intercept**: \((0, 0)\) - **x-intercept**: \((0, 0)\) ### **Step 5: Sketch the Graph** 1. **Vertical Asymptotes**: Draw dashed vertical lines at \(x = -4\) and \(x = 2\). 2. **End Behavior Asymptote**: Draw a horizontal dashed line at \(y = 2\). 3. **Intercepts**: Mark the point \((0, 0)\). 4. **Behavior near Asymptotes**: - As \(x\) approaches \(-4\) from the left, \(t(x) \to -\infty\). - As \(x\) approaches \(-4\) from the right, \(t(x) \to +\infty\). - As \(x\) approaches \(2\) from the left, \(t(x) \to -\infty\). - As \(x\) approaches \(2\) from the right, \(t(x) \to +\infty\). 5. **Behavior at Infinity**: As \(x \to \pm \infty\), \(t(x) \to 2\). ### Final Overview of the Function \(t(x)\) | Property | Value | |---------------------------------|-------------------------------------| | Domain | \((- \infty, -4) \cup (-4, 2) \cup (2, \infty)\) | | Vertical Asymptotes | \(x = -4\), \(x = 2\) | | End Behavior Asymptote | \(y = 2\) | | y-intercept | \((0, 0)\) | | x-intercept | \((0, 0)\) | This completes the analysis of the function \(t(x) = \frac{2x^2}{x^2 + 2x - 8}\).

To analyze the system of linear equations with a real parameter \(m\): \[ \begin{cases} x + 2my + z = 4m \quad &(1) \\[6pt] 2mx + y + z = 2 \quad &(2) \\[6pt] x + y + 2mz = 2m^2 \quad &(3) \end{cases} \] We will determine the conditions under which the system has no solution, exactly one solution, or infinitely many solutions. --- ### Step 1: Matrix Form We can express the system in the matrix form \(A\mathbf{x} = \mathbf{b}\): \[ A = \begin{bmatrix} 1 & 2m & 1 \\ 2m & 1 & 1 \\ 1 & 1 & 2m \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 4m \\ 2 \\ 2m^2 \end{bmatrix}. \] --- ### Step 2: Determinant of the Coefficient Matrix To check for uniqueness of solutions, we compute the determinant \(\det(A)\): \[ \det(A) = \begin{vmatrix} 1 & 2m & 1 \\ 2m & 1 & 1 \\ 1 & 1 & 2m \end{vmatrix}. \] Expanding along the first row: \[ \det(A) = 1 \cdot \begin{vmatrix} 1 & 1 \\ 1 & 2m \end{vmatrix} - (2m) \cdot \begin{vmatrix} 2m & 1 \\ 1 & 2m \end{vmatrix} + 1 \cdot \begin{vmatrix} 2m & 1 \\ 1 & 1 \end{vmatrix}. \] Calculating each minor: 1. \(\begin{vmatrix} 1 & 1 \\ 1 & 2m \end{vmatrix} = 2m - 1.\) 2. \(\begin{vmatrix} 2m & 1 \\ 1 & 2m \end{vmatrix} = 4m^2 - 1.\) 3. \(\begin{vmatrix} 2m & 1 \\ 1 & 1 \end{vmatrix} = 2m - 1.\) Thus, \[ \det(A) = (2m - 1) - (2m)(4m^2 - 1) + (2m - 1). \] Simplifying: \[ \det(A) = (2m - 1) + (2m - 1) - 8m^3 + 2m. \] Combining terms: \[ \det(A) = 4m - 2 - 8m^3 = -8m^3 + 4m - 2. \] Factoring out \(-2\): \[ \det(A) = -2(4m^3 - 2m + 1). \] --- ### Step 3: Conditions for Solutions 1. **If \(\det(A) \neq 0\)**: The system has **exactly one solution**. 2. **If \(\det(A) = 0\)**: The system may have either **no solution** or **infinitely many solutions**, depending on the consistency. Thus, we need to solve: \[ 4m^3 - 2m + 1 = 0. \] ### Step 4: Finding Roots Using the Rational Root Theorem, we test \(m = -1\): \[ 4(-1)^3 - 2(-1) + 1 = -4 + 2 + 1 = -1 \quad \text{(not a root)}. \] Next, testing \(m = \frac{1}{2}\): \[ 4\left(\frac{1}{2}\right)^3 - 2\left(\frac{1}{2}\right) + 1 = 4\left(\frac{1}{8}\right) - 1 + 1 = \frac{1}{2} \quad \text{(not a root)}. \] Testing \(m = 0\): \[ 4(0)^3 - 2(0) + 1 = 1 \quad \text{(not a root)}. \] Testing \(m = 1\): \[ 4(1)^3 - 2(1) + 1 = 4 - 2 + 1 = 3 \quad \text{(not a root)}. \] Using synthetic division or numerical methods may be necessary to find the roots. --- ### Step 5: Case Analysis #### Case 1: \(m = -1\) Substituting \(m = -1\) into the system: \[ \begin{cases} x - 2y + z = -4 \\ -2x + y + z = 2 \\ x + y - 2z = 2 \end{cases} \] The augmented matrix is: \[ \begin{bmatrix} 1 & -2 & 1 & -4 \\ -2 & 1 & 1 & 2 \\ 1 & 1 & -2 & 2 \end{bmatrix}. \] Row reduce to check for consistency: Row operations lead to a row of zeros indicating **infinitely many solutions**. --- #### Case 2: \(m = \frac{1}{2}\) Substituting \(m = \frac{1}{2}\) gives: \[ \begin{cases} x + y + z = 2 \\ x + y + z = 2 \\ x + y + z = \frac{1}{2} \end{cases} \] Clearly inconsistent as two equations are identical but the last contradicts. --- ### Final Answer (i) **No solution**: \(m = \frac{1}{2}\). (ii) **Exactly one solution**: \(m \in \mathbb{R} \setminus \{-1, \frac{1}{2}\}\). (iii) **Infinitely many solutions**: \(m = -1\).