To plot the graph of the given function f(x, y) = 2x^2 + 6xy + 9y^2 + 8x + 14, we can analyze the equation to identify its general shape and then find the local extremum points.
First, let's analyze the equation:
f(x, y) = 2x^2 + 6xy + 9y^2 + 8x + 14
The equation represents a quadratic function in two variables, x and y. Since the coefficients of the quadratic terms (x^2 and y^2) are both positive, the function represents an upward-opening paraboloid. The coefficients of the xy term (6xy) indicate that the paraboloid is symmetric about the x and y axes.
To find the local extremum points, we need to calculate the partial derivatives of the function with respect to x and y and solve the resulting system of equations to find the critical points.
Taking the partial derivative with respect to x:
∂f/∂x = 4x + 6y + 8
Taking the partial derivative with respect to y:
∂f/∂y = 6x + 18y
Setting both partial derivatives equal to zero, we have the following system of equations:
4x + 6y + 8 = 0 (Equation 1)
6x + 18y = 0 (Equation 2)
From Equation 2, we can simplify it to:
x + 3y = 0 (Equation 3)
Substituting Equation 3 into Equation 1, we have:
4(-3y) + 6y + 8 = 0
-12y + 6y + 8 = 0
-6y + 8 = 0
-6y = -8
y = 8/6
y = 4/3
Substituting the value of y into Equation 3, we can solve for x:
x + 3(4/3) = 0
x + 4 = 0
x = -4
Therefore, the critical point is (x, y) = (-4, 4/3).
To plot the graph, we can use a graphing software or manually create a 3D plot. The graph will show an upward-opening paraboloid centered around the origin. The local extremum point will be at (-4, 4/3), representing the minimum point on the paraboloid.