# A quadratic polynomial?

Hello, I just wanna clear up a confusion.

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

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

## Answers and Replies

Mark44
Mentor
Hello, I just wanna clear up a confusion.

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

where x and y are variables and a,b,c,d are constants.
No. A quadratic polynomial in two variables looks like this:
f(x, y) = ax2 + bxy + cy2 + dx + ey + f. See http://en.wikipedia.org/wiki/Quadratic_polynomial (two variables case).

HallsofIvy
Homework Helper
Mark44, I don't understand your response. You say "no" but then you give the general form for a quadratic in two variables which fits the one given by n0ya.
$a+ bx+ cy+ dxy$
is the same as $ax^2+ bxy+ cy^2+ dx+ ey+ f$.

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.

mathman
Hello, I just wanna clear up a confusion.

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

where x and y are variables and a,b,c,d are constants.
The usual terminology would call this bilinear.

Mark44
Mentor
I was under the impression that a and c could not both be zero, but that b could be zero. Clearly, we can't have a = b = c = 0, or the function wouldn't be quadratic, so there need to be some restrictions on a, b, and c (but not on d, e, and f). Perhaps the restriction is that not all of a, b, c can be zero.

Hello, I just wanna clear up a confusion.

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

where x and y are variables and a,b,c,d are constants.
To reduce this problem to its essence, take f(x,y) = xy. On the one hand there is no x2 or y2 term; but there is an xy term. So is this a quadratic or not?

chiro
Hey n0ya and welcome to the forums.

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).

My problem is really whether f(x,y) = xy is a quadratic polynomial, as SteveL27 points out.

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.

HallsofIvy
Homework Helper
Yes, it is a quadratic polynomial. If you were to rotate the coordinate system through 45 degrees, calling the new coordinates x' and y', f(x,y)= xy would become $x'^2- y'^2= 1$. Another way of looking at it is that the graph is a conic section, a hyperbola.

HallsofIvy
Homework Helper
Hello, I just wanna clear up a confusion.

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

where x and y are variables and a,b,c,d are constants.
The usual terminology would call this bilinear.
No, it would not. "xy" is not linear.

No, it would not. "xy" is not linear.
It's bilinear, meaning that

x(y1 + y2) = xy1 + xy2

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

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

mathman
In my experience there is a distinction between polynomials in one variable and polynomials in two variables. Highest order term x2y2 is called biquadratic and x3y3 is called bicubic.

Thanks for the replies.

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.

coolul007
Gold Member
Thanks for the replies.

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

Mark44
Mentor
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).
most of the time when you have an xy term the curve is a hyperbola
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.

coolul007
Gold Member
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.
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?

Mentallic
Homework Helper
BTW, where's the third dimension variable?
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=mx2, hence why it's called a hyperbolic paraboloid.

coolul007
Gold Member
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=mx2, hence why it's called a hyperbolic paraboloid.
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?

Mentallic
Homework Helper
I don't know. There seems to be quite a few different proposals and arguments for each term. I don't think we should settle on anything yet :tongue: