[imath]f(1, 2) = 2(1)^2 + 6(1)(2) +(2)^2 - 18 =2 + 12 + 4 - 18 = 0[/imath]I have to show that (1,2) is a solution to f(x,y)
But I get y equal to zero, ain't I supposed to find the stationary point?
So I'm just supposed to plugin (1,2) that seems odd considering other problems we're doing.[imath]f(1, 2) = 2(1)^2 + 6(1)(2) +(2)^2 - 18 =2 + 12 + 4 - 18 = 0[/imath]
It's a value on f(x,y). I don't know what you mean by "(1, 2) is a solution to f(x, y)."
Now, (1, 2) solves [imath]2x^2 + 6xy + y^2 = 18[/imath].
The stationary point of f(x, y) is (0, 0) as you have shown but that isn't part of your problem statement. Is what you posted the whole problem?
-Dan
That's all you need to do, then.So I'm just supposed to plugin (1,2) that seems odd considering other problems we're doing.
The problem as you cite it is stated incorrectly; f(x,y) is not an equation!I'm supposed to show that point (x_0,y_0)=(1,2) is a solution to the equation f(x,y)
"Vis, at punktet (?0,?0)=(1,2) er en løsning til ligningen." (translated)