Talk:Quadratic function

From Wikipedia, the free encyclopedia

True or false: If the ordinate is zero, then the axis is diagonal. (This came from what it says that depends on whether the ordinate is positive or negative.)

 -- If the ordinate is zero then the function looks like an X. It is not a hyperobola, it degenerates into two lines. I dont know if one could consider the "axis" to be both horizontal and vertical and thus undefined.

Contents 1 Why _quad_ratic? 2 Section ordering. 3 Matrix formulation 4 Java render

[edit] Why _quad_ratic?

Silly question perhaps, but why are they called quadratic? Highest power is _two_, number of terms is _three_; where's the four come from?

It's the same reason that x² is called "x-squared". Back when geometry was all of mathematics, a common problem people wanted to solve was quadrature, i.e. turning things into squares. Algebraically, problems involving squares and turning things into squares are always second power (because the area of a square with side x is x²). So call x² "x-squared" because it is the square associated with x, and call any equation involving this squared quantity a quadratic equation. Similarly, third degree functions are called cubic, rather than "ternary" or some other such -Lethe | Talk 18:06, Aug 22, 2004 (UTC)

[edit] Section ordering.

Would it be an a good idea to move the 'Roots' section below the 'Graph' section in the body.

I think the article would flow better that way and since the 'Graph' section contains the first part of the derivation of the equation for the roots as well. (As I remember them).

I'm asumming the derivation is not spelt out to stop people just copy the page for thier homework. ;-).

I think we should add a new article that discusses quadratic factoring. We also need better organization with this article because the sections are very randomly ordered. After organizing this article, let's add a paragraph about all the methods of factoring. Then we can provide a link to the new article (about factoring), which will go into the whole schmellalagang of factoring in detail. Anyone with me on this?

[edit] Matrix formulation

I think you should mention the matrix formulation of multivariate quadratics. Your formula is equivilant to , is a symetric 2 by 2 matrix, and are 2-vectors, and c is a scalar. I think the vertex is where the gradient () is zero: so , which can be solved easily by Cramers rule. The hessian matrix is everywhere , which is the shape operator of the plot-surface at the vertex. The quadratic can be rotated by a givens rotation to make into a diagonal matrix (call it ) and put the quadric in a standard orientation. I think the elements of are half the the principle curvatures of the plot at the vertex. I would add to the page directly, but I dont have time to double-check my facts first. The page probably should list the fundamental forms (I think they turn out to be very simple for quadratics).

[edit] Java render

In package java.awt.geom, there are classes QuadCurve2D, QuadCurve2D.Double, QuadCurve2D.Float. We can use them to draw a quadratic curve. In order to construct such an object, we need two points on the curve, and one control point. What does this control point mean? Jackzhp 19:09, 27 December 2006 (UTC)

Doesn't the documentation say what it means? But it probably has the obvious meaning: the line from this point to either of the other points is a tangent to the curve at that point. See the stuff on quadratic curves in the Bézier curve article. --Zundark 10:03, 28 December 2006 (UTC)

JDK's documentation doesn't say anything about the control point. However, your information gives the way to understand the stuff. Thanks. Furthermore, if we know y=ax^2+bx+c goes through point A & B, we need a formula to get the control point C. Conversely, if we know the point A,B, & C, we should have a formula to find a,b,& c. Can somebody post the fomulas here?Jackzhp 01:42, 30 December 2006 (UTC)

What makes you think the curve is given by y = ax² + bx + c? The way you describe the classes, the axis of the parabola needn't be vertical. --Zundark 18:47, 30 December 2006 (UTC)

Thanks for your responding. ok. suppose the curve is ax^2+bxy+cy^2+dx+ey+f=0 goes through point A(xA,yA) & B(xB,yB). we need a formula to get the control point C(xC,yC). then we can use these three points to draw the segment between point A&B. What is the formula? Conversely, if we know the points A,B,C, we need the formula to find a,b,c,d,e,f. What is this formula again? Thanks. Jackzhp 21:11, 31 December 2006 (UTC)