What is the extreme P(x,y)=x^2-y^2+6y-9 ?

The extreme P(x,y)=x^2-y^2+6y-9 is Saddle(0,3)

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extreme P(x,y)=x^2-y^2+6y-9

extreme points

P (x, y) = x^{2} - y^{2} + 6 y - 9

Saddle (0, 3)

Solution steps

Find the critical points:

(0, 3)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 2

D (x, y) = - 4

Check the sign of at each critical point

Check critical point

(0, 3) :

Saddle

Saddle (0, 3)

f (x, y) = \frac{x ^{3}}{3} - y^{2} - 5 x + x y - 3

h (x) = - 2 x^{2} - 1, - 4 \leq x \leq 3

x^{3} - 12 x^{2} + 45 x - 4

y^{4} - y^{2} - 2 x y + x^{2}

f (x, y) = lo g_{10} (\frac{( x ^{2} + 2 \cdot y )}{( x + y ) ^{2}})

What is the extreme P(x,y)=x^2-y^2+6y-9 ?
The extreme P(x,y)=x^2-y^2+6y-9 is Saddle(0,3)