Point on f(x,y)

hajfajv

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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?
 

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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?
[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
 
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)
 
[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
So I'm just supposed to plugin (1,2) that seems odd considering other problems we're doing.
 
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)
The problem as you cite it is stated incorrectly; f(x,y) is not an equation!

But it appears that you invented the function f(x,y) as part of your work, and the problem actually said,

Show that (1,2) is a solution of the equation​
1651666898024.png

This is why we ask for an exact copy of the entire problem!

But substitution is all you need to do to show that a given ordered pair satisfies (is a solution of) an equation. Sometimes easy problems are inserted among harder ones to remind you of the basics. If you showed us the other problems, we might be able to see the reason.
 
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