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Is f(x,y) = a + bx + cy + dxy

a quadratic polynomial?

where x and y are variables and a,b,c,d are constants.

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- #1

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Is f(x,y) = a + bx + cy + dxy

a quadratic polynomial?

where x and y are variables and a,b,c,d are constants.

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Mark44

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No. A quadratic polynomial in two variables looks like this:

Is f(x,y) = a + bx + cy + dxy

a quadratic polynomial?

where x and y are variables and a,b,c,d are constants.

f(x, y) = ax

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HallsofIvy

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[itex]a+ bx+ cy+ dxy[/itex]

is the same as [itex]ax^2+ bxy+ cy^2+ dx+ ey+ f[/itex].

Those are the same with your a= 0, b equal to nOya's d, c= 0, d equal to n0ya's b, e equal to n0ya's c, and your f equal to n0ya's a.

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mathman

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The usual terminology would call this bilinear.

Is f(x,y) = a + bx + cy + dxy

a quadratic polynomial?

where x and y are variables and a,b,c,d are constants.

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Mark44

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To reduce this problem to its essence, take f(x,y) = xy. On the one hand there is no x

Is f(x,y) = a + bx + cy + dxy

a quadratic polynomial?

where x and y are variables and a,b,c,d are constants.

- #7

chiro

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As a guiding rule for further questions like this, look at the highest term (in terms of the order) and use that as a basis to figure out what kind of equation/polynomial/blah something is on top of other things that need to be considered (for example polynomials only have integer powers so a x^(1/2) in the expression would make it a non-polynomial instantly).

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I know that this polynomial is of degree 2, and according to Wikipedia, "a quadratic polynomial or quadratic is a polynomial of degree two", it is a quadratic polynomial. However, some argue that a x^2 or a y^2 term is needed in order to call it 'quadratic', so I just wanted to clear up this confusion.

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HallsofIvy

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HallsofIvy

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Is f(x,y) = a + bx + cy + dxy

a quadratic polynomial?

where x and y are variables and a,b,c,d are constants.

No, it would not. "xy" is not linear.The usual terminology would call this bilinear.

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It's bilinear, meaning thatNo, it would not. "xy" is not linear.

x(y1 + y2) = xy1 + xy2

and x(ay) = a(xy) for a constant a;

and likewise in y. It's linear in each variable.

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mathman

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Then, as far as I understand it, if you for example have: x = t; y = t, then f(x,y)=xy will be a parabola (t^2).

Anyway, I think both quadratic and bilinear are both correct terms here. However, it might be more intuative to call it a bilinear mapping, instead of a quadratic polynomial, since we have no x^2 or y^2 terms.

- #14

coolul007

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most of the time when you have an xy term the curve is a hyperbola

Then, as far as I understand it, if you for example have: x = t; y = t, then f(x,y)=xy will be a parabola (t^2).

Anyway, I think both quadratic and bilinear are both correct terms here. However, it might be more intuative to call it a bilinear mapping, instead of a quadratic polynomial, since we have no x^2 or y^2 terms.

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Mark44

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Then, as far as I understand it, if you for example have: x = t; y = t, then f(x,y)=xy will be a parabola (t^2).

f(x, y) = xy represents a surface in three dimensions, not a curve in the plane, as are parabolas and hyperbolas. According to wolframalpha, the surface is a hyperbolic paraboloid.most of the time when you have an xy term the curve is a hyperbola

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coolul007

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I think you found a bug, if you plot the old fashioned way n=xy, you get all of the real factors of n and it's a hyperbola. BTW, where's the third dimension variable?f(x, y) = xy represents a surface in three dimensions, not a curve in the plane, as are parabolas and hyperbolas. According to wolframalpha, the surface is a hyperbolic paraboloid.

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Mentallic

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f(x,y) IS the 3rd dimension variable. It's basically equivalent to z=xy. For any fixed value z, then yes, you're dealing with a hyperbola, but if you consider the graph intersected by the plane y=mx, then you're dealing with a parabola z=mxBTW, where's the third dimension variable?

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coolul007

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It must be too early, I was thinking f(x) not f(x,y)...so what is the answer to the original question? did we settle on bi-linear?f(x,y) IS the 3rd dimension variable. It's basically equivalent to z=xy. For any fixed value z, then yes, you're dealing with a hyperbola, but if you consider the graph intersected by the plane y=mx, then you're dealing with a parabola z=mx^{2}, hence why it's called a hyperbolic paraboloid.

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Mentallic

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http://mathworld.wolfram.com/BilinearFunction.html...so what is the answer to the original question? did we settle on bi-linear?

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