Write the formula for Newton's method and use the given initial approximation to compute the approximations \( x_{1} \) and \( x_{2} \). \( f(x)=x^{2}+2 x-15, x_{0}=4 \) B. \( x_{n+1}=x_{n}+\frac{x_{n}^{2}+2 x_{n}-15}{2 x_{n}+2} \) C. \( x_{n+1}=x_{n}-\frac{2 x_{n}+2}{x_{n}^{2}+2 x_{n}-15} \) D. \( x_{n+1}=x_{n}-\frac{x_{n}^{2}+2 x_{n}-15}{2 x_{n}+2} \) Use the given initial approximation to compute the approximations \( x_{1} \) and \( x_{2} \). \( x_{1}=\square \) (Do not round until the final answer. Then round to three decimal places as needed.) \( x_{2}=\square \) (Do not round until the final answer. Then round to three decimal places as needed.) (D)

To apply Newton's Method to the function \( f(x) = x^{2} + 2x - 15 \) with the initial approximation \( x_{0} = 4 \), follow these steps: ### Newton's Method Formula Newton's Method is given by the iterative formula: \[ x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})} \] where: - \( f(x) = x^{2} + 2x - 15 \) - \( f'(x) = 2x + 2 \) Thus, the specific update formula for this function is: \[ x_{n+1} = x_{n} - \frac{x_{n}^{2} + 2x_{n} - 15}{2x_{n} + 2} \] ### Computing \( x_{1} \) Start with the initial approximation \( x_{0} = 4 \): \[ x_{1} = 4 - \frac{4^{2} + 2 \cdot 4 - 15}{2 \cdot 4 + 2} \] Calculate the numerator and denominator separately: \[ \text{Numerator} = 4^{2} + 2 \cdot 4 - 15 = 16 + 8 - 15 = 9 \] \[ \text{Denominator} = 2 \cdot 4 + 2 = 8 + 2 = 10 \] Now, compute \( x_{1} \): \[ x_{1} = 4 - \frac{9}{10} = 4 - 0.9 = 3.1 \] ### Computing \( x_{2} \) Using \( x_{1} = 3.1 \): \[ x_{2} = 3.1 - \frac{3.1^{2} + 2 \cdot 3.1 - 15}{2 \cdot 3.1 + 2} \] Calculate the numerator and denominator: \[ \text{Numerator} = 3.1^{2} + 2 \cdot 3.1 - 15 = 9.61 + 6.2 - 15 = 0.81 \] \[ \text{Denominator} = 2 \cdot 3.1 + 2 = 6.2 + 2 = 8.2 \] Now, compute \( x_{2} \): \[ x_{2} = 3.1 - \frac{0.81}{8.2} \approx 3.1 - 0.09878 \approx 3.0012 \] Rounding to three decimal places: \[ x_{2} \approx 3.001 \] ### Final Approximations \[ x_{1} = 3.1 \] \[ x_{2} = 3.001 \]