quadratic

Let f(x) = ax² + bx + c.

 

\(\begin{cases}a + b + c = 2\\4a + 2b + c = -7\\16a + 4b + c = 13\end{cases}\)

 

Solving by matrix method(or whatever method you like):

 

\(\begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}1&1&1\\4&2&1\\16&4&1\end{bmatrix}^{-1}\cdot \begin{bmatrix}2\\-7\\13\end{bmatrix} = \begin{bmatrix}\dfrac{19}3\\-28\\\dfrac{71}3\end{bmatrix} \)

 

Therefore \(f(x) = ax^2 + bx + c = \dfrac{19}3 x^2 - 28x + \dfrac{71}3\)

 

Substituting x = 5 gives the answer. I believe you can do it from here.