Suppose experimental data are represented by a set of points in the plane. An interpolating polynomial for the data is a polynomial whose graph passes through every point. In scientific work, such a polynomial can be used, for example, to estimate values between the known data points. Another use is to create curves for graphical images on a computer screen. One method for finding an interpolating polynomial is to solve a system of linear equations. Find the interpolating polynomial \( p(t)=a_{0}+a_{1} t+a_{2} t^{2} \) for the data \( (1,11),(2,19),(3,23) \). That is, find \( a_{0}, a_{1} \), and \( a_{2} \) such that the following is true. \[ \begin{array}{l} a_{0}+a_{1}(1)+a_{2}(1)^{2}=11 \\ a_{0}+a_{1}(2)+a_{2}(2)^{2}=19 \\ a_{0}+a_{1}(3)+a_{2}(3)^{2}=23 \end{array} \] Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The interpolating polynomial is \( p(t)= \) \( \square \) . B. There are infinitely many possible interpolating polynomials. C. There does not exist an interpolating polynomial for the given data.
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To find the interpolating polynomial \( p(t) = a_{0} + a_{1} t + a_{2} t^{2} \), we need to solve the given system of equations: 1. \( a_{0} + a_{1} \cdot 1 + a_{2} \cdot 1^2 = 11 \) 2. \( a_{0} + a_{1} \cdot 2 + a_{2} \cdot 2^2 = 19 \) 3. \( a_{0} + a_{1} \cdot 3 + a_{2} \cdot 3^2 = 23 \) This translates to: 1. \( a_{0} + a_{1} + a_{2} = 11 \) 2. \( a_{0} + 2a_{1} + 4a_{2} = 19 \) 3. \( a_{0} + 3a_{1} + 9a_{2} = 23 \) Next, we can eliminate \( a_{0} \) from these equations. Subtract equation 1 from equations 2 and 3: From equations (2 - 1): \[ (2a_{1} + 4a_{2}) - (a_{1} + a_{2}) = 19 - 11 \] \[ a_{1} + 3a_{2} = 8 \] → (4) From equations (3 - 1): \[ (3a_{1} + 9a_{2}) - (a_{1} + a_{2}) = 23 - 11 \] \[ 2a_{1} + 8a_{2} = 12 \] Dividing this equation by 2: \[ a_{1} + 4a_{2} = 6 \] → (5) Now we have: 1. \( a_{1} + 3a_{2} = 8 \) (equation 4) 2. \( a_{1} + 4a_{2} = 6 \) (equation 5) Subtract equation 4 from equation 5: \[ (a_{1} + 4a_{2}) - (a_{1} + 3a_{2}) = 6 - 8 \] \[ a_{2} = -2 \] Now substitute \( a_{2} \) back into equation 4: \[ a_{1} + 3(-2) = 8 \] \[ a_{1} - 6 = 8 \] \[ a_{1} = 14 \] Now substitute \( a_{1} \) and \( a_{2} \) back into equation 1 to find \( a_{0} \): \[ a_{0} + 14 - 2 = 11 \] \[ a_{0} + 12 = 11 \] \[ a_{0} = -1 \] Thus, we have: - \( a_{0} = -1 \) - \( a_{1} = 14 \) - \( a_{2} = -2 \) The interpolating polynomial is: \[ p(t) = -1 + 14t - 2t^{2} \] So, the correct choice is: A. The interpolating polynomial is \( p(t) = -2t^2 + 14t - 1 \).
