A function \( f \) and a point \( P \) are given. Let \( \theta \) correspond to the direction of the directional derivative. Complete parts (a) through (e). \( f(x, y)=16-3 x^{2}-2 y^{2}, P(2,3) \) a. Find the gradient and evaluate it at \( P \). The gradient at \( P \) is \( \langle\square \).

To find the gradient of the function \( f(x, y) = 16 - 3x^2 - 2y^2 \), we first compute the partial derivatives. 1. The partial derivative with respect to \( x \) is: \[ f_x = \frac{\partial}{\partial x}(16 - 3x^2 - 2y^2) = -6x \] 2. The partial derivative with respect to \( y \) is: \[ f_y = \frac{\partial}{\partial y}(16 - 3x^2 - 2y^2) = -4y \] Now we can evaluate these partial derivatives at the point \( P(2, 3) \): - For \( x = 2 \): \[ f_x(2, 3) = -6(2) = -12 \] - For \( y = 3 \): \[ f_y(2, 3) = -4(3) = -12 \] Thus, the gradient at the point \( P(2, 3) \) is: \[ \nabla f(P) = \langle -12, -12 \rangle \] The gradient at \( P \) is \( \langle -12, -12 \rangle \).