Examine the function for relative extrema and saddle points \[ h(x, y)=x^{2}-7 x y-y^{2} \]

To find the relative extrema and saddle points of the function \( h(x, y) = x^2 - 7xy - y^2 \), we first need to find the critical points. We do this by computing the first partial derivatives and setting them equal to zero: \[ h_x = \frac{\partial h}{\partial x} = 2x - 7y, \] \[ h_y = \frac{\partial h}{\partial y} = -7x - 2y. \] Setting \( h_x = 0 \) and \( h_y = 0 \) gives us the system of equations: 1. \( 2x - 7y = 0 \) 2. \( -7x - 2y = 0 \) From equation (1), we can express \( y = \frac{2}{7}x \). Substituting this into equation (2): \[ -7x - 2\left(\frac{2}{7}x\right) = 0 \implies -7x - \frac{4}{7}x = 0 \implies -\frac{49x + 4x}{7} = 0 \implies -53x = 0. \] This implies \( x = 0 \). Plugging \( x = 0 \) back into the equation for \( y \) gives \( y = 0 \). Therefore, the critical point is \( (0, 0) \). Next, we use the second derivative test to classify this critical point. We need to compute the second partial derivatives: \[ h_{xx} = \frac{\partial^2 h}{\partial x^2} = 2, \] \[ h_{yy} = \frac{\partial^2 h}{\partial y^2} = -2, \] \[ h_{xy} = \frac{\partial^2 h}{\partial x \partial y} = -7. \] Now, we compute the determinant of the Hessian matrix: \[ D = h_{xx}h_{yy} - (h_{xy})^2 = (2)(-2) - (-7)^2 = -4 - 49 = -53. \] Since \( D < 0 \), we conclude that the critical point \( (0, 0) \) is a saddle point. To summarize, the function \( h(x, y) = x^2 - 7xy - y^2 \) has a saddle point at \( (0, 0) \), which means the surface defined by the function curves up in some directions and down in others around this point. It’s a definite "whoa" moment when you visualize how that saddle point behaves!