REFER TO IMAGE ## Problem StatementFind and classify the critical points of \( f(x, y) = (y - 2)x^2 - y^2 \). Justify your answers.### Solution OutlineTo find the critical points of the function \( f(x, y) = (y - 2)x^2 - y^2 \), we need to follow these steps:1. **Find the Partial Derivatives:** Determine the partial derivatives of the function with respect to \( x \) and \( y \). 2. **Set Partial Derivatives to Zero:** Set these partial derivatives equal to zero to find the critical points. 3. **Second Partial Derivative Test:** Use the second partial derivative test to classify the nature of each critical point.### Step 1: Find the Partial DerivativesLet \( f(x, y) = (y - 2)x^2 - y^2 \).**Partial derivative with respect to \( x \):**\( f_x = \frac{\partial}{\partial x} [(y - 2)x^2 - y^2] \)\( f_x = (y - 2) \cdot 2x \)\( f_x = 2x(y - 2) \)**Partial derivative with respect to \( y \):**\( f_y = \frac{\partial}{\partial y} [(y - 2)x^2 - y^2] \)\( f_y = x^2 - 2y \)### Step 2: Set Partial Derivatives to ZeroSet the partial derivatives equal to zero to find the critical points.**For \( f_x = 0 \):**\( 2x(y - 2) = 0 \)\( x = 0 \quad \text{or} \quad y = 2 \)**For \( f_y = 0 \):**\( x^2 - 2y = 0 \)### Solve Simultaneously1. **Case 1:** \( x = 0 \)\[ x^2 - 2y = 0 \implies 0 - 2y = 0 \implies y = 0 \]So, one critical point is \( (0, 0) \).2. **Case 2:** \( y = 2 \)\[ x^2 - 2(2

Name: REFER TO IMAGE ## Problem StatementFind and classify the critical poin
Uploaded: 2024-03-20T07:07:37.000Z
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Description: REFER TO IMAGE ## Problem StatementFind and classify the critical points of \( f(x, y) = (y - 2)x^2 - y^2 \). Justify your answers.### Solution OutlineTo find the critical points of the function \( f(x, y) = (y - 2)x^2 - y^2 \), we need to follow the