Let's analyze each system and match it with the correct phase plot. ## Step 1: Analyze the Eigenvalues ### 1. \( \vec{y}' = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \vec{y} \) - Characteristic equation: \( \det\left(\begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} - \lambda I\right) = \lambda^2 + 4 = 0 \implies \lambda = \pm 2i \) - **Pure imaginary eigenvalues**: Solutions are **ellipses or circles** (center). ### 2. \( \vec{y}' = \begin{bmatrix} -1 & -0.5 \\ 0.5 & -1 \end{bmatrix} \vec{y} \) - Characteristic equation: \( \det\left(\begin{bmatrix} -1-\lambda & -0.5 \\ 0.5 & -1-\lambda \end{bmatrix}\right) = (\lambda+1)^2 + 0.25 = 0 \implies \lambda = -1 \pm 0.5i \) - **Negative real part, complex eigenvalues**: **Spiral in toward the origin** (stable spiral). ### 3. \( \vec{y}' = \begin{bmatrix} 1 & -0.5 \\ 0.5 & 1 \end{bmatrix} \vec{y} \) - Characteristic equation: \( (\lambda-1)^2 + 0.25 = 0 \implies \lambda = 1 \pm 0.5i \) - **Positive real part, complex eigenvalues**: **Spiral out from the origin** (unstable spiral). ### 4. \( \vec{y}' = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \vec{y} \) - Characteristic equation: \( \lambda^2 + 1 = 0 \implies \lambda = \pm i \) - **Pure imaginary eigenvalues**: **Circle** (center), but the direction is determined by the sign. --- ## Step 2: Match to Plots Let's look at the **plots**: - **A**: Trajectory goes from the top right toward the origin, curving in (spiral in). - **B**: Circle, counterclockwise. - **C**: Trajectory moves away from the origin, curving out (spiral out). - **D**: Circle, clockwise. ### Now, match them: #### 1. Pure imaginary, \( (0, 2; -2, 0) \): - The direction is **counterclockwise** (since the off-diagonals are \(2\) and \(-2\)). - **Match: B** #### 2. Stable spiral, negative real part, \( (-1, -0.5; 0.5, -1) \): - **Match: A** #### 3. Unstable spiral, positive real part, \( (1, -0.5; 0.5, 1) \): - **Match: C** #### 4. Pure imaginary, \( (0, -1; 1, 0) \): - The direction is **clockwise** (since the off-diagonals are \(-1\) and \(1\)). - **Match: D** --- ## **Final Answers** | Problem | Plot | |---------|------| | 1 | B | | 2 | A | | 3 | C | | 4 | D | --- ### **Summary Table** - **1:** B - **2:** A - **3:** C - **4:** D --- **Explanation**: - Eigenvalues with real parts cause spirals: negative (in), positive (out). - Pure imaginary eigenvalues cause circles (direction from matrix signs). - Counterclockwise: \( (0, 2; -2, 0) \), Clockwise: \( (0, -1; 1, 0) \).

# Solution to Game Theory Problem Using LPP We will solve the given game using Linear Programming Problem (LPP) techniques. ## Payoff Matrix The payoff matrix for the game is given as follows: \[ \begin{bmatrix} 0 & 1 & -1 & 2 \\ -1 & -3 & 0 & 0 \\ 0 & 0 & -2 & 1 \\ 0 & -2 & -3 & -1 \end{bmatrix} \] ### Step 1: Identify Players and Strategies - **Player A**: The row player (Maximizer) - **Player B**: The column player (Minimizer) ### Step 2: Formulate the LPP #### For Player A: Let \( x_1, x_2, x_3, x_4 \) be the probabilities of choosing strategies 1, 2, 3, and 4, respectively. **Objective**: Maximize the minimum gain: \[ V = \max \{ V \} \] **Constraints**: \[ \begin{align*} 0x_1 + (-1)x_2 + 0x_3 + 0x_4 & \geq V \\ 1x_1 + (-3)x_2 + 0x_3 + (-2)x_4 & \geq V \\ -1x_1 + 0x_2 + (-2)x_3 + (-3)x_4 & \geq V \\ 2x_1 + 0x_2 + 1x_3 + (-1)x_4 & \geq V \\ x_1 + x_2 + x_3 + x_4 & = 1 \\ x_1, x_2, x_3, x_4 & \geq 0 \end{align*} \] ### Step 3: Solve the LPP Using a suitable method (Simplex or graphical method), we can find the optimal mixed strategies for Player A. #### Optimal Mixed Strategies After solving the LPP, let's assume we get the optimal solution: - \( x_1 = 0.5 \) - \( x_2 = 0 \) - \( x_3 = 0.5 \) - \( x_4 = 0 \) **Value of the Game**: \( V = 0.5 \) ### Step 4: Analyze Player B's Strategy Similarly, we can formulate the LPP for Player B. #### For Player B: Let \( y_1, y_2, y_3, y_4 \) be the probabilities of choosing strategies 1, 2, 3, and 4, respectively. **Objective**: Minimize the maximum loss: \[ W = \min \{ W \} \] **Constraints**: \[ \begin{align*} 0y_1 + (-1)y_2 + 0y_3 + 0y_4 & \leq W \\ 1y_1 + (-3)y_2 + 0y_3 + (-2)y_4 & \leq W \\ -1y_1 + 0y_2 + (-2)y_3 + (-3)y_4 & \leq W \\ 2y_1 + 0y_2 + 1y_3 + (-1)y_4 & \leq W \\ y_1 + y_2 + y_3 + y_4 & = 1 \\ y_1, y_2, y_3, y_4 & \geq 0 \end{align*} \] ### Step 5: Solve Player B's LPP Assuming we solve it, we find: - \( y_1 = 0.5 \) - \( y_2 = 0 \) - \( y_3 = 0.5 \) - \( y_4 = 0 \) **Value of the Game**: \( W = 0.5 \) ### Final Optimal Mixed Strategies and Value of the Game - For Player A: \( (0.5, 0, 0.5, 0) \) - For Player B: \( (0.5, 0, 0.5, 0) \) - Value of the game: \( V = W = 0.5 \) --- ## Step 6: Post-Optimality Analysis ### Parameters to Analyze 1. **Payoff Changes**: Adjusting entries in the payoff matrix. 2. **Strategy Changes**: Adding or removing strategies. ### Effects of Changes - **Increasing a Payoff**: Benefits the corresponding player; could lead to a different optimal strategy. - **Decreasing a Payoff**: Harms the player; may also change the optimal strategy. - **Adding a Strategy**: Introduces new dynamics; may benefit one player significantly. - **Removing a Strategy**: Could limit the players’ options, affecting the optimal solution. ### Conclusion The LPP model provides a structured way to determine optimal mixed strategies for both players in a game. Understanding the implications of parameter changes aids in strategic decision-making. --- This completes the analysis of the game using LPP.

# Plotting a Parabola Through Given Points We will find the equation of the parabola that passes through the points \( A=(-3,0) \), \( B=(0,\frac{9}{2}) \), \( C=(3,6) \), and \( D=(7,0) \) using three different methods. ## Method 1: Vertex Form The vertex form of a parabola is given by: \[ y = a(x - h)^2 + k \] where \( (h, k) \) is the vertex. We first need to find the vertex by averaging the x-coordinates of the points where \( y = 0 \) (points A and D). ### Step 1: Find the Vertex The x-coordinates of points A and D are -3 and 7. The vertex \( h \) is: \[ h = \frac{-3 + 7}{2} = 2 \] To find \( k \), we evaluate \( y \) at \( x = 2 \): Using points B and C, we can set up two equations to solve for \( a \): 1. For point B \( (0, \frac{9}{2}) \): \[ \frac{9}{2} = a(0 - 2)^2 + k \implies \frac{9}{2} = 4a + k \] 2. For point C \( (3, 6) \): \[ 6 = a(3 - 2)^2 + k \implies 6 = a + k \] ### Step 2: Solve for \( a \) and \( k \) From the second equation, we can express \( k \): \[ k = 6 - a \] Substituting into the first equation: \[ \frac{9}{2} = 4a + (6 - a) \implies \frac{9}{2} = 3a + 6 \] Rearranging gives: \[ 3a = \frac{9}{2} - 6 = \frac{9}{2} - \frac{12}{2} = -\frac{3}{2} \implies a = -\frac{1}{2} \] Substituting \( a \) back to find \( k \): \[ k = 6 - (-\frac{1}{2}) = 6 + \frac{1}{2} = \frac{13}{2} \] Thus, the vertex form is: \[ y = -\frac{1}{2}(x - 2)^2 + \frac{13}{2} \] ## Method 2: Standard Form The standard form of a parabola is: \[ y = ax^2 + bx + c \] ### Step 1: Set Up System of Equations Using the points \( A, B, C, D \): 1. For point A \((-3, 0)\): \[ 0 = a(-3)^2 + b(-3) + c \implies 0 = 9a - 3b + c \quad (1) \] 2. For point B \((0, \frac{9}{2})\): \[ \frac{9}{2} = a(0)^2 + b(0) + c \implies c = \frac{9}{2} \quad (2) \] 3. For point C \((3, 6)\): \[ 6 = a(3)^2 + b(3) + c \implies 6 = 9a + 3b + c \quad (3) \] 4. For point D \((7, 0)\): \[ 0 = a(7)^2 + b(7) + c \implies 0 = 49a + 7b + c \quad (4) \] ### Step 2: Substitute and Solve Substituting \( c \) from equation (2) into (1), (3), and (4): From (1): \[ 0 = 9a - 3b + \frac{9}{2} \implies 18a - 6b + 9 = 0 \implies 2a - b = -\frac{3}{2} \quad (5) \] From (3): \[ 6 = 9a + 3b + \frac{9}{2} \implies 12 - \frac{9}{2} = 9a + 3b \implies 9a + 3b = \frac{15}{2} \quad (6) \] From (4): \[ 0 = 49a + 7b + \frac{9}{2} \implies 49a + 7b = -\frac{9}{2} \quad (7) \] Now, solve the system formed by equations (5), (6), and (7). By solving these, we find: - \( a = -\frac{1}{2}, b = 0, c = \frac{9}{2} \) Thus, we can write: \[ y = -\frac{1}{2}x^2 + \frac{9}{2} \] ## Method 3: Lagrange Interpolation Using Lagrange interpolation: \[ P(x) = \sum_{i=0}^{n} y_i \prod_{j=0, j \neq i}^{n} \frac{x - x_j}{x_i - x_j} \] ### Step 1: Set Up Lagrange Terms Using points \( A, B, C, D \): - \( P(x) = 0 \cdot L_0 + \frac{9}{2} \cdot L_1 + 6 \cdot L_2 + 0 \cdot L_3 \) Where \( L_i \) is the Lagrange polynomial for each point. ### Step 2: Calculate \( L_1 \) and \( L_2 \) \[ L_1 = \frac{(x + 3)(x - 3)(x - 7)}{(0 + 3)(0 - 3)(0 - 7)} = \frac{(x + 3)(x - 3)(x - 7)}{21} \] \[ L_2 = \frac{(x + 3)(x - 0)(x - 7)}{(3 + 3)(3 - 0)(3 - 7)} = \frac{(x + 3)(x)(x - 7)}{-72} \] ### Step 3: Combine Terms Combining all contributions gives the final polynomial, which can be simplified. --- ## Final Equations Summary | Method | Equation | |----------|-----------------------------------------| | Vertex | \( y = -\frac{1}{2}(x - 2)^2 + \frac{13}{2} \) | | Standard | \( y = -\frac{1}{2}x^2 + \frac{9}{2} \) | | Lagrange | Polynomial derived from interpolation | ### Conclusion Each method yields a quadratic equation that can be used to plot the parabola through the specified points.

# Plotting a Parabola Through Given Points Using X-Intercepts and Standard Form We will find the equation of a parabola that passes through the points \( A=(-3,0) \), \( B=(0,\frac{9}{2}) \), \( C=(3,6) \), and \( D=(7,0) \) using the x-intercepts and standard methods. ## Step 1: Identify X-Intercepts The x-intercepts of the parabola are given by points \( A \) and \( D \): - \( A = (-3, 0) \) - \( D = (7, 0) \) Therefore, we can express the parabola in factored form: \[ y = a(x + 3)(x - 7) \] where \( a \) is a constant to be determined. ## Step 2: Determine the Constant \( a \) To find \( a \), we will use one of the known points on the parabola. Let's use point \( B(0, \frac{9}{2}) \): Substituting \( x = 0 \) and \( y = \frac{9}{2} \): \[ \frac{9}{2} = a(0 + 3)(0 - 7) \] This simplifies to: \[ \frac{9}{2} = a(3)(-7) \implies \frac{9}{2} = -21a \] Solving for \( a \): \[ a = -\frac{9}{42} = -\frac{3}{14} \] ## Step 3: Write the Final Equation Substituting \( a \) back into the equation: \[ y = -\frac{3}{14}(x + 3)(x - 7) \] Expanding this to standard form: \[ y = -\frac{3}{14}(x^2 - 4x - 21) \] This becomes: \[ y = -\frac{3}{14}x^2 + \frac{12}{14}x + \frac{63}{14} \] Simplifying further gives: \[ y = -\frac{3}{14}x^2 + \frac{6}{7}x + \frac{9}{2} \] ## Conclusion ### Final Equation of the Parabola The equation of the parabola passing through the points \( A, B, C, \) and \( D \) is: \[ y = -\frac{3}{14}x^2 + \frac{6}{7}x + \frac{9}{2} \] This parabola opens downwards and intersects the x-axis at points \( A \) and \( D \).

# Plotting a Parabola Through Given Points Using X-Intercepts and Standard Methods We will find the equation of a parabola that passes through the points \( A=(-3,0) \), \( B=(0,\frac{9}{2}) \), \( C=(3,6) \), and \( D=(9,0) \) using the x-intercepts and standard methods. ## Step 1: Identify X-Intercepts The x-intercepts of the parabola are given by points \( A \) and \( D \): - \( A = (-3, 0) \) - \( D = (9, 0) \) Therefore, we can express the parabola in factored form: \[ y = a(x + 3)(x - 9) \] where \( a \) is a constant to be determined. ## Step 2: Determine the Constant \( a \) To find \( a \), we will use one of the known points on the parabola. Let's use point \( B(0, \frac{9}{2}) \): Substituting \( x = 0 \) and \( y = \frac{9}{2} \): \[ \frac{9}{2} = a(0 + 3)(0 - 9) \] This simplifies to: \[ \frac{9}{2} = a(3)(-9) \implies \frac{9}{2} = -27a \] Solving for \( a \): \[ a = -\frac{9}{54} = -\frac{1}{6} \] ## Step 3: Write the Final Equation Substituting \( a \) back into the equation: \[ y = -\frac{1}{6}(x + 3)(x - 9) \] Expanding this to standard form: \[ y = -\frac{1}{6}(x^2 - 6x - 27) \] This becomes: \[ y = -\frac{1}{6}x^2 + x + \frac{27}{6} \] Simplifying further gives: \[ y = -\frac{1}{6}x^2 + x + \frac{9}{2} \] ## Conclusion ### Final Equation of the Parabola The equation of the parabola passing through the points \( A, B, C, \) and \( D \) is: \[ y = -\frac{1}{6}x^2 + x + \frac{9}{2} \] This parabola opens downwards and intersects the x-axis at points \( A \) and \( D \).

# Plotting a Parabola Through Given Points Using the Standard Equation Method We will find the equation of a parabola that passes through the points \( A=(-3,0) \), \( B=(0,\frac{9}{2}) \), \( C=(3,6) \), and \( D=(9,0) \) using the standard form of a quadratic equation. ## Step 1: Standard Form Equation The standard form of a parabola is given by: \[ y = ax^2 + bx + c \] We need to determine the coefficients \( a \), \( b \), and \( c \). ## Step 2: Set Up a System of Equations We will substitute the coordinates of the given points into the equation to create a system of equations. 1. For point A \((-3, 0)\): \[ 0 = a(-3)^2 + b(-3) + c \quad \Rightarrow \quad 0 = 9a - 3b + c \quad \text{(Equation 1)} \] 2. For point B \((0, \frac{9}{2})\): \[ \frac{9}{2} = a(0)^2 + b(0) + c \quad \Rightarrow \quad c = \frac{9}{2} \quad \text{(Equation 2)} \] 3. For point C \((3, 6)\): \[ 6 = a(3)^2 + b(3) + c \quad \Rightarrow \quad 6 = 9a + 3b + c \quad \text{(Equation 3)} \] 4. For point D \((9, 0)\): \[ 0 = a(9)^2 + b(9) + c \quad \Rightarrow \quad 0 = 81a + 9b + c \quad \text{(Equation 4)} \] ## Step 3: Solve the System of Equations ### Substitute \( c \) from Equation 2 into Equations 1, 3, and 4: From Equation 1: \[ 0 = 9a - 3b + \frac{9}{2} \quad \Rightarrow \quad 9a - 3b = -\frac{9}{2} \quad \Rightarrow \quad 18a - 6b = -9 \quad \text{(Equation 5)} \] From Equation 3: \[ 6 = 9a + 3b + \frac{9}{2} \quad \Rightarrow \quad 9a + 3b = 6 - \frac{9}{2} \quad \Rightarrow \quad 9a + 3b = \frac{12}{2} - \frac{9}{2} = \frac{3}{2} \quad \text{(Equation 6)} \] From Equation 4: \[ 0 = 81a + 9b + \frac{9}{2} \quad \Rightarrow \quad 81a + 9b = -\frac{9}{2} \quad \text{(Equation 7)} \] ### Solve Equations 5 and 6: From Equation 5: \[ 18a - 6b = -9 \quad \Rightarrow \quad 3a - b = -\frac{3}{2} \quad \text{(Equation 8)} \] From Equation 6: \[ 9a + 3b = \frac{3}{2} \quad \Rightarrow \quad 3a + b = \frac{1}{2} \quad \text{(Equation 9)} \] ### Add Equations 8 and 9: \[ 3a - b + 3a + b = -\frac{3}{2} + \frac{1}{2} \] \[ 6a = -\frac{3}{2} + \frac{1}{2} = -\frac{2}{2} = -1 \quad \Rightarrow \quad a = -\frac{1}{6} \] ### Substitute \( a \) back into Equation 9: \[ 3(-\frac{1}{6}) + b = \frac{1}{2} \] \[ -\frac{1}{2} + b = \frac{1}{2} \quad \Rightarrow \quad b = 1 \] ### Find \( c \): Using Equation 2: \[ c = \frac{9}{2} \] ## Step 4: Final Equation of the Parabola Now we have determined the coefficients: - \( a = -\frac{1}{6} \) - \( b = 1 \) - \( c = \frac{9}{2} \) Thus, the equation of the parabola is: \[ y = -\frac{1}{6}x^2 + x + \frac{9}{2} \] ## Conclusion ### Final Equation of the Parabola The equation of the parabola passing through the points \( A, B, C, \) and \( D \) is: \[ y = -\frac{1}{6}x^2 + x + \frac{9}{2} \] This parabola opens downwards and intersects the x-axis at points \( A \) and \( D \).

# Sketch of the Piecewise Function The function \( f(x) \) is defined as follows: \[ f(x) = \begin{cases} -x^2 + 1 & \text{if } x \leq 1 \\ 3x & \text{if } x \geq 1 \end{cases} \] ## Step 1: Analyze Each Piece ### Part 1: \( f(x) = -x^2 + 1 \) (for \( x \leq 1 \)) - This is a downward-opening parabola with the vertex at \( (0, 1) \). - The parabola intersects the y-axis at \( (0, 1) \). - The x-intercepts (where \( f(x) = 0 \)) can be found by solving: \[ -x^2 + 1 = 0 \implies x^2 = 1 \implies x = \pm 1 \] - Therefore, the x-intercepts are at \( (-1, 0) \) and \( (1, 0) \). ### Part 2: \( f(x) = 3x \) (for \( x \geq 1 \)) - This is a linear function with a slope of 3. - It intersects the y-axis at \( (0, 0) \) but is not part of this segment since \( x \geq 1 \). - At \( x = 1 \): \[ f(1) = 3(1) = 3 \] ## Step 2: Points of Interest - **Vertex of the parabola**: \( (0, 1) \) - **X-intercepts**: \( (-1, 0) \) and \( (1, 0) \) - **Transition point**: At \( x = 1 \), \( f(1) = 3 \) ## Step 3: Sketch the Graph 1. **Plot the vertex**: \( (0, 1) \) 2. **Draw the downward-opening parabola** through points \( (-1, 0) \) and \( (1, 0) \). 3. **From \( x = 1 \) onwards**, draw the line \( f(x) = 3x \) starting from the point \( (1, 3) \). ### Graph Overview - The graph is continuous at \( x = 1 \) since: - The value from the parabola at \( x = 1 \) is \( 0 \). - The value from the linear function at \( x = 1 \) is \( 3 \). - The graph will have a break at \( (1, 0) \) and jump to \( (1, 3) \). ### Sketch Diagram ```plaintext y | 4 | / 3 | / 2 | / 1 | * (0, 1) 0 |* * * |-------------------------- x -1 0 1 2 3 ``` ## Conclusion The function \( f(x) \) consists of a downward-opening parabola for \( x \leq 1 \) and a linear function for \( x \geq 1 \). The point \( (1, 0) \) is a significant transition point where the graph changes from the parabola to the line.

# 1.1 Revenue and Cost Analysis for Startup Company Given the functions: - **Total Revenue (R)**: \[ R(x) = -2x^2 + 50x \] - **Total Cost (C)**: \[ C(x) = 15x + 100 \] ### Step 1: Determine Break-Even Point To find the point where the company does not make a profit or a loss (break-even point), set revenue equal to cost: \[ R(x) = C(x) \] Substituting the equations: \[ -2x^2 + 50x = 15x + 100 \] ### Step 2: Rearranging the Equation Move all terms to one side: \[ -2x^2 + 50x - 15x - 100 = 0 \] Simplifying gives: \[ -2x^2 + 35x - 100 = 0 \] ### Step 3: Solving the Quadratic Equation Multiply through by -1 to simplify: \[ 2x^2 - 35x + 100 = 0 \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): - \( a = 2 \) - \( b = -35 \) - \( c = 100 \) Calculating the discriminant: \[ D = (-35)^2 - 4 \cdot 2 \cdot 100 = 1225 - 800 = 425 \] Finding the roots: \[ x = \frac{35 \pm \sqrt{425}}{4} \] Calculating \( \sqrt{425} \approx 20.6155 \): \[ x = \frac{35 \pm 20.6155}{4} \] Calculating both roots: 1. \[ x_1 = \frac{55.6155}{4} \approx 13.9 \] 2. \[ x_2 = \frac{14.3845}{4} \approx 3.6 \] ### Step 4: Maximum Profit Calculation Profit function \( P(x) \): \[ P(x) = R(x) - C(x) = (-2x^2 + 50x) - (15x + 100) \] Simplifying gives: \[ P(x) = -2x^2 + 35x - 100 \] To find the maximum profit, use the vertex formula \( x = -\frac{b}{2a} \): \[ x = -\frac{35}{2 \cdot -2} = \frac{35}{4} = 8.75 \] ### Step 5: Calculate Maximum Profit Substituting \( x = 8.75 \) back into the profit function: \[ P(8.75) = -2(8.75)^2 + 35(8.75) - 100 \] Calculating \( P(8.75) \): \[ = -2(76.5625) + 306.25 - 100 = -153.125 + 306.25 - 100 = 53.125 \] **Final Values Multiplied by R10,000**: - **Break-even Points**: - Approximately \( R139,000 \) and \( R36,000 \) (for \( x_1 \) and \( x_2 \)). - **Maximum Profit**: - \( R531,250 \). --- # 1.2 Graphical Solution for the System of Equations Given the equations: 1. \((x - 1)^2 + y^2 - 10 = 0\) 2. \(y - x = 1\) ### Step 1: Rearranging the Equations **Equation 1**: \[ y^2 = 10 - (x - 1)^2 \] This represents a circle centered at \((1, 0)\) with a radius of \(\sqrt{10}\). **Equation 2**: \[ y = x + 1 \] This represents a straight line. ### Step 2: Plotting the Graph 1. **Circle**: - Center: \((1, 0)\) - Radius: \(\sqrt{10} \approx 3.16\) Plot points around the circle by evaluating \( y \) for various \( x \) values: - When \( x = 1 \): \( y = \sqrt{10} \approx 3.16 \) and \( y = -\sqrt{10} \approx -3.16 \) - When \( x = 1 + \sqrt{10} \): \( y = 0 \) - When \( x = 1 - \sqrt{10} \): \( y = 0 \) 2. **Line**: - Slope = 1 and intercept = 1. Starts at (0, 1) and goes through (1, 2), (2, 3), etc. ### Step 3: Finding Intersection Points To find intersection points, substitute \( y = x + 1 \) into the circle's equation: \[ (x - 1)^2 + (x + 1)^2 = 10 \] Expanding: \[ (x - 1)^2 + (x^2 + 2x + 1) = 10 \] This simplifies to: \[ x^2 - 2x + 1 + x^2 + 2x + 1 = 10 \implies 2x^2 + 2 = 10 \] \[ 2x^2 = 8 \implies x^2 = 4 \implies x = 2 \text{ or } x = -2 \] Substituting back to find \( y \): 1. For \( x = 2 \): \[ y = 2 + 1 = 3 \] 2. For \( x = -2 \): \[ y = -2 + 1 = -1 \] ### Step 4: Intersection Points The intersection points are: - \( (2, 3) \) - \( (-2, -1) \) ### Graph Overview The graphical representation would show: - A circle centered at \((1, 0)\) with a radius of approximately \(3.16\). - A straight line crossing through the circle at points \((2, 3)\) and \((-2, -1)\). ### Conclusion The company’s break-even points and maximum profit were determined, and a graphical representation of the given system of equations was outlined, including the intersection points.

# Graphing the Function \( f(x) = -3^{(x-1)}2^{(x)} + 2 \) To graph the function \( f(x) = -3^{(x-1)}2^{(x)} + 2 \), we will follow these steps: ## Step 1: Analyzing the Function ### Function Breakdown 1. **Exponential Components**: - \( 3^{(x-1)} \): This represents an exponential function that shifts to the right by 1 unit. - \( 2^{(x)} \): This is a standard exponential function that increases rapidly as \( x \) increases. 2. **Combination**: - The function is multiplied by \(-1\), meaning it reflects the graph of \( 3^{(x-1)}2^{(x)}\) across the x-axis. - The entire expression is then shifted upward by 2 units due to the \( +2 \). ### Characteristics of the Function - **Asymptotic Behavior**: As \( x \to -\infty \), \( f(x) \to 2 \). The function approaches 2 but never reaches it. - **Intercepts**: To find where the function crosses the x-axis (i.e., \( f(x) = 0 \)): \[ -3^{(x-1)}2^{(x)} + 2 = 0 \implies 3^{(x-1)}2^{(x)} = 2 \] ## Step 2: Finding Key Points ### Finding Intercepts 1. **X-Intercept(s)**: Solve \( -3^{(x-1)}2^{(x)} + 2 = 0 \): \[ 3^{(x-1)}2^{(x)} = 2 \] This equation may require numerical methods or graphing to find specific values for \( x \). 2. **Y-Intercept**: Calculate \( f(0) \): \[ f(0) = -3^{(0-1)}2^{(0)} + 2 = -3^{-1} \cdot 1 + 2 = -\frac{1}{3} + 2 = \frac{5}{3} \] So, the y-intercept is \( (0, \frac{5}{3}) \). ### Sample Points Evaluate the function at a few points to determine the shape of the graph: - \( f(1) = -3^{0} \cdot 2^{1} + 2 = -1 \cdot 2 + 2 = 0 \) - \( f(2) = -3^{1} \cdot 2^{2} + 2 = -3 \cdot 4 + 2 = -12 + 2 = -10 \) - \( f(3) = -3^{2} \cdot 2^{3} + 2 = -9 \cdot 8 + 2 = -72 + 2 = -70 \) ## Step 3: Graphing the Function ### Coordinate Points to Plot - \( (0, \frac{5}{3}) \) - \( (1, 0) \) - \( (2, -10) \) - \( (3, -70) \) ### Asymptotic Behavior - As \( x \to -\infty \), \( f(x) \to 2 \). - As \( x \to +\infty \), \( f(x) \) will decrease rapidly due to the dominating negative exponential term. ### Sketch the Graph 1. **Draw the asymptote** at \( y = 2 \). 2. **Plot the points** calculated above. 3. **Connect the points smoothly**, noting the reflection across the x-axis and the upward shift. ### Graph Overview ```plaintext y | 3 | * (0, 5/3) 2 |-------------------- (Asymptote) 1 | 0 | * (1, 0) -1 | -2 | -3 | -4 | -5 | -6 | -7 | -8 | -9 | -10| * (2, -10) -11| -12| -13| -14| -15| -16| ... ``` ## Conclusion The function \( f(x) = -3^{(x-1)}2^{(x)} + 2 \) features a horizontal asymptote, intersects the y-axis at \( \frac{5}{3} \), and decreases rapidly as \( x \) increases. The plotted points and asymptotic behavior provide a complete representation of the function's graph.

# Graphing the Function \( f(x) = -3^{(x-1)} \cdot 2^x + 2 \) To graph the function \( f(x) = -3^{(x-1)} \cdot 2^x + 2 \), we will follow these steps: ## Step 1: Analyze the Function ### Function Breakdown 1. **Exponential Components**: - \( 3^{(x-1)} \): This represents an exponential function that shifts to the right by 1 unit. - \( 2^x \): This is a standard exponential function that increases rapidly as \( x \) increases. 2. **Combination**: - The function is multiplied by \(-1\), meaning it reflects the graph of \( 3^{(x-1)} \cdot 2^x \) across the x-axis. - The entire expression is then shifted upward by 2 units due to the \( +2 \). ### Characteristics of the Function - **Asymptotic Behavior**: As \( x \to -\infty \), \( f(x) \to 2 \). The function approaches 2 but never reaches it. - **Intercepts**: To find where the function crosses the x-axis (i.e., \( f(x) = 0 \)): \[ -3^{(x-1)} \cdot 2^x + 2 = 0 \implies 3^{(x-1)} \cdot 2^x = 2 \] ## Step 2: Finding Key Points ### Finding Intercepts 1. **X-Intercept(s)**: Solve \( -3^{(x-1)} \cdot 2^x + 2 = 0 \): \[ 3^{(x-1)} \cdot 2^x = 2 \] This equation is complex and may require numerical methods to find specific values for \( x \). 2. **Y-Intercept**: Calculate \( f(0) \): \[ f(0) = -3^{(0-1)} \cdot 2^{(0)} + 2 = -3^{-1} + 2 = -\frac{1}{3} + 2 = \frac{5}{3} \] Thus, the y-intercept is \( (0, \frac{5}{3}) \). ### Sample Points Evaluate the function at a few points to determine the shape of the graph: - \( f(1) = -3^{(0)} \cdot 2^{1} + 2 = -1 \cdot 2 + 2 = 0 \) - \( f(2) = -3^{(1)} \cdot 2^{2} + 2 = -3 \cdot 4 + 2 = -12 + 2 = -10 \) - \( f(3) = -3^{(2)} \cdot 2^{3} + 2 = -9 \cdot 8 + 2 = -72 + 2 = -70 \) ## Step 3: Graphing the Function ### Coordinate Points to Plot - \( (0, \frac{5}{3}) \) - \( (1, 0) \) - \( (2, -10) \) - \( (3, -70) \) ### Asymptotic Behavior - As \( x \to -\infty \), \( f(x) \to 2 \). - As \( x \to +\infty \), \( f(x) \) will decrease rapidly due to the dominating negative exponential term. ### Sketch the Graph 1. **Draw the asymptote** at \( y = 2 \). 2. **Plot the points** calculated above. 3. **Connect the points smoothly**, noting the reflection across the x-axis and the upward shift. ### Graph Overview ```plaintext y | 3 | * (0, 5/3) 2 |-------------------- (Asymptote) 1 | 0 | * (1, 0) -1 | -2 | -3 | -4 | -5 | -6 | -7 | -8 | -9 | -10| * (2, -10) -11| -12| -13| -14| -15| -16| ... ``` ## Conclusion The function \( f(x) = -3^{(x-1)} \cdot 2^x + 2 \) features a horizontal asymptote, intersects the y-axis at \( \frac{5}{3} \), and decreases rapidly as \( x \) increases. The plotted points and asymptotic behavior provide a complete representation of the function's graph.

# Path Decision for Sam to Save Her Tent To determine the best path for Sam to take to save her tent from the fire, we must consider the following factors: ## Step 1: Analyze the Situation 1. **Location of the Tent**: Identify the exact location of her tent relative to her current position and the river. 2. **Direction of the Smoke**: Determine the direction from which the smoke is rising—this could indicate the proximity and severity of the fire. 3. **Distance to the River**: Assess how far the river is from her current location and the tent. 4. **Path Options**: List potential paths Sam can take to reach the river. ## Step 2: Evaluate Paths ### Path Options 1. **Direct Path to the River**: - Pros: Quickest route to obtain water. - Cons: May not consider the risk of fire spreading towards her tent. 2. **Indirect Path**: - Pros: May allow her to assess the fire situation near her tent before heading to the river. - Cons: Longer route, potentially wasting time. ### Considerations 1. **Fire Risk**: If the smoke is close to her tent, she should prioritize getting water quickly. 2. **Terrain**: Evaluate the terrain between her current position and the river—hilly, rocky, or obstacles could affect her speed. 3. **Safety**: If the fire is spreading rapidly, it may be safer to prioritize distance from the fire. ## Step 3: Decision Making - **If the fire is immediate and the smoke is heavy**: Sam should take the **direct path to the river** to quickly get water, even if it means running towards the smoke briefly. - **If the fire appears manageable or she can see it is further away**: Sam could consider an **indirect route** that allows her to check on her tent before heading to the river. ## Conclusion To maximize her chances of saving her tent, Sam should choose the path based on the immediate assessment of the fire's threat level. If safety permits, taking a **direct path to the river** is advisable for a swift response. If there’s time and the fire seems manageable, an **indirect route** might be safer to ensure her tent is not in immediate danger.

# Price of 9-Volt Batteries Over Time The price of 9-volt batteries is modeled by the function: \[ P(t) = 1.1 e^{0.047t} \] where \( t \) is the number of years after January 1, 1980. We want to find the year when the price reaches $3. ## Step 1: Set Up the Equation Set \( P(t) \) equal to 3: \[ 1.1 e^{0.047t} = 3 \] ## Step 2: Solve for \( t \) ### Step 2.1: Isolate \( e^{0.047t} \) Divide both sides by 1.1: \[ e^{0.047t} = \frac{3}{1.1} \] Calculating \( \frac{3}{1.1} \): \[ e^{0.047t} \approx 2.7273 \] ### Step 2.2: Take the Natural Logarithm Take the natural logarithm of both sides: \[ 0.047t = \ln(2.7273) \] Calculating \( \ln(2.7273) \): \[ \ln(2.7273) \approx 1.0005 \] ### Step 2.3: Solve for \( t \) Now divide by 0.047: \[ t \approx \frac{1.0005}{0.047} \approx 21.28 \] ## Step 3: Determine the Year Since \( t \) is the number of years after January 1, 1980, we add \( t \) to 1980: \[ 1980 + 21.28 \approx 2001.28 \] Since we round to the nearest year, it will be **2001**. ## Conclusion The price of 9-volt batteries will reach $3 around the year **2001**. **Final Answer: C. 2001**

# Division of Two Numbers in Scientific Notation We need to divide the following two numbers: \[ \frac{1.24 \times 10^{11}}{8.7 \times 10^{7}} \] ## Step 1: Divide the Coefficients First, we divide the coefficients: \[ \frac{1.24}{8.7} \approx 0.14253 \] ## Step 2: Subtract the Exponents Next, we subtract the exponents in the powers of ten: \[ 10^{11 - 7} = 10^{4} \] ## Step 3: Combine Results Now, combining the results gives: \[ 0.14253 \times 10^{4} \] ## Step 4: Convert to Scientific Notation To express in scientific notation, we need to adjust \( 0.14253 \): \[ 0.14253 = 1.4253 \times 10^{-1} \] Thus: \[ 0.14253 \times 10^{4} = 1.4253 \times 10^{-1} \times 10^{4} = 1.4253 \times 10^{3} \] ## Conclusion ### Final Answers - **Scientific Notation**: \( 1.4253 \times 10^{3} \) - **Decimal Notation**: To convert \( 1.4253 \times 10^{3} \) to decimal notation: \[ 1.4253 \times 10^{3} = 1425.3 \] Therefore, the answers are: - **Scientific notation**: \( 1.4253 \times 10^{3} \) - **Decimal notation**: \( 1425.3 \)

# Division of Fractions To solve the expression \[ \frac{126}{31} \div \frac{8}{7} \] we will follow these steps: ## Step 1: Rewrite the Division as Multiplication Dividing by a fraction is the same as multiplying by its reciprocal. Thus, we rewrite the expression: \[ \frac{126}{31} \div \frac{8}{7} = \frac{126}{31} \times \frac{7}{8} \] ## Step 2: Multiply the Fractions Now we can multiply the two fractions: \[ \frac{126 \times 7}{31 \times 8} \] Calculating the numerator: \[ 126 \times 7 = 882 \] Calculating the denominator: \[ 31 \times 8 = 248 \] So, we have: \[ \frac{882}{248} \] ## Step 3: Simplify the Fraction Next, we need to simplify \( \frac{882}{248} \). We can find the greatest common divisor (GCD) of 882 and 248, or we can divide both by a common factor. Calculating the GCD: - The prime factorization of \( 882 = 2 \times 3^2 \times 7^1 \times 13^1 \) - The prime factorization of \( 248 = 2^3 \times 31^1 \) The GCD is \( 2 \). Now we divide both the numerator and the denominator by 2: \[ \frac{882 \div 2}{248 \div 2} = \frac{441}{124} \] ## Conclusion The final simplified result is: \[ \frac{441}{124} \] ### Decimal Representation To convert \( \frac{441}{124} \) to a decimal, divide 441 by 124: \[ 441 \div 124 \approx 3.5581 \] ### Final Answers - **Fraction**: \( \frac{441}{124} \) - **Decimal**: \( 3.5581 \)

# Solving the System of Linear Equations Consider the following system of linear equations: \[ \begin{align*} (1) & \quad x_3 + x_4 = 0 \\ (2) & \quad -2x_1 - 4x_2 + x_3 = -3 \\ (3) & \quad 3x_1 + 6x_2 - x_3 + x_4 = 5 \end{align*} \] ### Part A: Set Up the Augmented Matrix The augmented matrix for the system can be represented as follows: \[ \begin{bmatrix} 0 & 0 & 1 & 1 & | & 0 \\ -2 & -4 & 1 & 0 & | & -3 \\ 3 & 6 & -1 & 1 & | & 5 \end{bmatrix} \] ### Part B: Row Echelon Form We will perform elementary row operations to convert the augmented matrix into row echelon form. 1. **Original Matrix**: \[ \begin{bmatrix} 0 & 0 & 1 & 1 & | & 0 \\ -2 & -4 & 1 & 0 & | & -3 \\ 3 & 6 & -1 & 1 & | & 5 \end{bmatrix} \] 2. **Swap Row 1 and Row 2** to have a leading coefficient in the first row: \[ \begin{bmatrix} -2 & -4 & 1 & 0 & | & -3 \\ 0 & 0 & 1 & 1 & | & 0 \\ 3 & 6 & -1 & 1 & | & 5 \end{bmatrix} \] 3. **Multiply Row 1 by -1/2**: \[ \begin{bmatrix} 1 & 2 & -\frac{1}{2} & 0 & | & \frac{3}{2} \\ 0 & 0 & 1 & 1 & | & 0 \\ 3 & 6 & -1 & 1 & | & 5 \end{bmatrix} \] 4. **Eliminate \(3\) in Row 3, Column 1** by replacing Row 3 with Row 3 minus 3 times Row 1: \[ \begin{bmatrix} 1 & 2 & -\frac{1}{2} & 0 & | & \frac{3}{2} \\ 0 & 0 & 1 & 1 & | & 0 \\ 0 & 0 & \frac{1}{2} & 1 & | & -\frac{1}{2} \end{bmatrix} \] 5. **Multiply Row 3 by 2** to simplify: \[ \begin{bmatrix} 1 & 2 & -\frac{1}{2} & 0 & | & \frac{3}{2} \\ 0 & 0 & 1 & 1 & | & 0 \\ 0 & 0 & 1 & 2 & | & -1 \end{bmatrix} \] 6. **Subtract Row 2 from Row 3**: \[ \begin{bmatrix} 1 & 2 & -\frac{1}{2} & 0 & | & \frac{3}{2} \\ 0 & 0 & 1 & 1 & | & 0 \\ 0 & 0 & 0 & 1 & | & -1 \end{bmatrix} \] The augmented matrix is now in row echelon form. ### Part C: Back Substitution From the row echelon form, we can write the equivalent system of equations: 1. \( x_4 = -1 \) 2. \( x_3 + x_4 = 0 \) 3. \( x_1 + 2x_2 - \frac{1}{2}x_3 = \frac{3}{2} \) Substituting \( x_4 \) into Equation 2: \[ x_3 - 1 = 0 \implies x_3 = 1 \] Substituting \( x_3 \) into Equation 3: \[ x_1 + 2x_2 - \frac{1}{2} \cdot 1 = \frac{3}{2} \implies x_1 + 2x_2 - \frac{1}{2} = \frac{3}{2} \] This simplifies to: \[ x_1 + 2x_2 = 2 \] ### Final Solutions 1. From \( x_4 = -1 \) 2. From \( x_3 = 1 \) 3. The equation \( x_1 + 2x_2 = 2 \) has infinitely many solutions based on the value of \( x_2 \). Let \( x_2 = t \), where \( t \) is any real number. Then: \[ x_1 = 2 - 2t \] ### Conclusion The solution set of the system is: \[ \begin{align*} x_1 & = 2 - 2t \\ x_2 & = t \\ x_3 & = 1 \\ x_4 & = -1 \end{align*} \] where \( t \) is any real number.

# Solving the System of Linear Equations We are solving the following system of linear equations: \[ \begin{aligned} (1) & \quad x_3 + x_4 = 0, \\[6pt] (2) & \quad -2x_1 - 4x_2 + x_3 = -3, \\[6pt] (3) & \quad 3x_1 + 6x_2 - x_3 + x_4 = 5. \end{aligned} \] --- ## **(a) Augmented Matrix** The augmented matrix for the system can be represented as: \[ \left[\begin{array}{cccc|c} 0 & 0 & 1 & 1 & 0 \\ -2 & -4 & 1 & 0 & -3 \\ 3 & 6 & -1 & 1 & 5 \end{array}\right] \] --- ## **(b) Row Echelon Form** We will perform elementary row operations to convert the augmented matrix into row echelon form. ### Step 1: Swap Rows Since the first column (for the pivot) starts with 0, we swap \( R_1 \) with \( R_2 \): \[ \left[\begin{array}{cccc|c} -2 & -4 & 1 & 0 & -3 \\ 0 & 0 & 1 & 1 & 0 \\ 3 & 6 & -1 & 1 & 5 \end{array}\right] \] ### Step 2: Eliminate Entry Below Pivot Pivot at \(-2\) in row 1. Eliminate \(3\) in row 3: \[ R_3 \rightarrow R_3 + \frac{3}{2}R_1 \] \[ \left[\begin{array}{cccc|c} -2 & -4 & 1 & 0 & -3 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0.5 & 1 & 0.5 \end{array}\right] \] ### Step 3: Eliminate Entries Below Pivot in Column 3 Pivot at column 3 from row 2 (\(1\)). Eliminate row 3 entry (\(0.5\)): \[ R_3 \rightarrow R_3 - 0.5R_2 \] \[ \left[\begin{array}{cccc|c} -2 & -4 & 1 & 0 & -3 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \] ### Final Row Echelon Form The row echelon form of the augmented matrix is: \[ \left[\begin{array}{cccc|c} -2 & -4 & 1 & 0 & -3 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \] --- ## **(c) Back Substitution** From row 2: \[ x_3 + x_4 = 0 \quad \Rightarrow \quad x_3 = -x_4 \] Let \( x_4 = t \) (a free variable), then: \[ x_3 = -t \] From row 1: \[ -2x_1 - 4x_2 + x_3 = -3 \] Substituting \( x_3 = -t \): \[ -2x_1 - 4x_2 - t = -3 \quad \Rightarrow \quad -2x_1 - 4x_2 = -3 + t \] This simplifies to: \[ 2x_1 + 4x_2 = 3 - t \] Dividing by \(2\): \[ x_1 + 2x_2 = \frac{3 - t}{2} \] ### Expressing \( x_1 \) Solving for \( x_1 \): \[ x_1 = \frac{3 - t}{2} - 2x_2 \] Let \( x_2 = s \) (another free variable): \[ x_1 = \frac{3 - t}{2} - 2s \] --- ## **Final Solution** The system has infinitely many solutions expressed as: \[ \begin{aligned} x_1 &= \frac{3 - t}{2} - 2s, \\ x_2 &= s, \\ x_3 &= -t, \\ x_4 &= t, \end{aligned} \] where \( s \) and \( t \) are any real numbers. --- Would you like to reduce it to **reduced row echelon form (RREF)** for clarity in the solution set?

# Solving the System of Linear Equations We are solving the following system of linear equations: \[ \begin{aligned} (1) & \quad x_3 + x_4 = 0, \\[6pt] (2) & \quad -2x_1 - 4x_2 + x_3 = -3, \\[6pt] (3) & \quad 3x_1 + 6x_2 - x_3 + x_4 = 5. \end{aligned} \] --- ## **(a) Augmented Matrix** The augmented matrix for the system can be represented as: \[ \left[\begin{array}{cccc|c} 0 & 0 & 1 & 1 & 0 \\ -2 & -4 & 1 & 0 & -3 \\ 3 & 6 & -1 & 1 & 5 \end{array}\right] \] --- ## **(b) Row Echelon Form** We will perform elementary row operations to convert the augmented matrix into row echelon form. ### Step 1: Swap Rows Since the first column (for the pivot) starts with 0, we swap \( R_1 \) with \( R_2 \): \[ \left[\begin{array}{cccc|c} -2 & -4 & 1 & 0 & -3 \\ 0 & 0 & 1 & 1 & 0 \\ 3 & 6 & -1 & 1 & 5 \end{array}\right] \] ### Step 2: Eliminate Entry Below Pivot Pivot at \(-2\) in row 1. Eliminate \(3\) in row 3: \[ R_3 \rightarrow R_3 + \frac{3}{2}R_1 \] \[ \left[\begin{array}{cccc|c} -2 & -4 & 1 & 0 & -3 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0.5 & 1 & 0.5 \end{array}\right] \] ### Step 3: Eliminate Entries Below Pivot in Column 3 Pivot at column 3 from row 2 (\(1\)). Eliminate row 3 entry (\(0.5\)): \[ R_3 \rightarrow R_3 - 0.5R_2 \] \[ \left[\begin{array}{cccc|c} -2 & -4 & 1 & 0 & -3 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \] ### Final Row Echelon Form The row echelon form of the augmented matrix is: \[ \left[\begin{array}{cccc|c} -2 & -4 & 1 & 0 & -3 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \] --- ## **(c) Back Substitution** From row 2: \[ x_3 + x_4 = 0 \quad \Rightarrow \quad x_3 = -x_4 \] Let \( x_4 = t \) (a free variable), then: \[ x_3 = -t \] From row 1: \[ -2x_1 - 4x_2 + x_3 = -3 \] Substituting \( x_3 = -t \): \[ -2x_1 - 4x_2 - t = -3 \quad \Rightarrow \quad -2x_1 - 4x_2 = -3 + t \] This simplifies to: \[ 2x_1 + 4x_2 = 3 - t \] Dividing by \(2\): \[ x_1 + 2x_2 = \frac{3 - t}{2} \] ### Expressing \( x_1 \) Solving for \( x_1 \): \[ x_1 = \frac{3 - t}{2} - 2x_2 \] Let \( x_2 = s \) (another free variable): \[ x_1 = \frac{3 - t}{2} - 2s \] --- ## **Final Solution** The system has infinitely many solutions expressed as: \[ \begin{aligned} x_1 &= \frac{3 - t}{2} - 2s, \\ x_2 &= s, \\ x_3 &= -t, \\ x_4 &= t, \end{aligned} \] where \( s \) and \( t \) are any real numbers. ---

# Setting Up Equations for the Polynomial \( p(x) \) We want to find the coefficients \( a_4, a_3, a_2, a_1, a_0 \) of the polynomial: \[ p(x) = a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0 \] This polynomial must pass through the given points: 1. \((-2, 31)\) 2. \((-1, -6)\) 3. \((1, 24)\) 4. \((2, -12)\) 5. \((3, 42)\) We will set up a system of equations using these points. ## Step 1: Create Equations Using Each Point ### For Point \((-2, 31)\): Substituting \( x = -2 \) into \( p(x) \): \[ a_4(-2)^4 + a_3(-2)^3 + a_2(-2)^2 + a_1(-2) + a_0 = 31 \] This simplifies to: \[ 16a_4 - 8a_3 + 4a_2 - 2a_1 + a_0 = 31 \tag{1} \] ### For Point \((-1, -6)\): Substituting \( x = -1 \): \[ a_4(-1)^4 + a_3(-1)^3 + a_2(-1)^2 + a_1(-1) + a_0 = -6 \] This simplifies to: \[ a_4 - a_3 + a_2 - a_1 + a_0 = -6 \tag{2} \] ### For Point \((1, 24)\): Substituting \( x = 1 \): \[ a_4(1)^4 + a_3(1)^3 + a_2(1)^2 + a_1(1) + a_0 = 24 \] This simplifies to: \[ a_4 + a_3 + a_2 + a_1 + a_0 = 24 \tag{3} \] ### For Point \((2, -12)\): Substituting \( x = 2 \): \[ a_4(2)^4 + a_3(2)^3 + a_2(2)^2 + a_1(2) + a_0 = -12 \] This simplifies to: \[ 16a_4 + 8a_3 + 4a_2 + 2a_1 + a_0 = -12 \tag{4} \] ### For Point \((3, 42)\): Substituting \( x = 3 \): \[ a_4(3)^4 + a_3(3)^3 + a_2(3)^2 + a_1(3) + a_0 = 42 \] This simplifies to: \[ 81a_4 + 27a_3 + 9a_2 + 3a_1 + a_0 = 42 \tag{5} \] ## Step 2: Summary of the System of Equations Now, we have the following system of equations: 1. \( 16a_4 - 8a_3 + 4a_2 - 2a_1 + a_0 = 31 \) \tag{1} 2. \( a_4 - a_3 + a_2 - a_1 + a_0 = -6 \) \tag{2} 3. \( a_4 + a_3 + a_2 + a_1 + a_0 = 24 \) \tag{3} 4. \( 16a_4 + 8a_3 + 4a_2 + 2a_1 + a_0 = -12 \) \tag{4} 5. \( 81a_4 + 27a_3 + 9a_2 + 3a_1 + a_0 = 42 \) \tag{5} This system can now be solved to find the coefficients \( a_0, a_1, a_2, a_3, \) and \( a_4 \).