# Solution ## Step 1: Find Level Curves for \( k = , 1, -1 \) Given: \[ f(x, y) = y^2 + x - 2 \] Level curves are found by setting \( f(x, y) = k \): \[ y^2 + x - 2 = k \implies x = k + 2 - y^2 \] - For \( k = \): \( x = 2 - y^2 \) - For \( k = 1 \): \( x = 3 - y^2 \) - For \( k = -1 \): \( x = 1 - y^2 \) **Explanation:** Each value of \( k \) gives a different parabola opening to the left on the \(xy\)-plane. --- ## Step 2: Find the Gradient Field The gradient of \( f(x, y) \) is: \[ \nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (1, 2y) \] So, at any point \((x, y)\), the gradient vector is \((1, 2y)\). **Explanation:** The gradient vector at each point points in the direction of greatest increase of \(f\), and its magnitude depends on \(y\). --- ## Final Answer - **Level curves**: Parabolas \( x = 2 - y^2,\, x = 3 - y^2,\, x = 1 - y^2 \) for \( k = , 1, -1 \). - **Gradient field**: At every point \((x, y)\), the vector is \((1, 2y)\). You would plot these parabolas and, at various points, draw arrows pointing in the direction \((1, 2y)\) to represent the gradient field.