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() = + is called the vertex form, where h and k are the x and y coordinates of the vertex, respectively. The coefficient a is the same value in all three forms. To convert the standard form to factored form , one needs only the quadratic formula to determine the two roots r 1 and r 2 .
As an example, x 2 + 5x + 6 factors as ... The real part is the x-coordinate of the vertex. ... as solutions to quadratic equations or as coefficients in an equation. ...
That is, h is the x-coordinate of the axis of symmetry (i.e. the axis of symmetry has equation x = h), and k is the minimum value (or maximum value, if a < 0) of the quadratic function. One way to see this is to note that the graph of the function f(x) = x 2 is a parabola whose vertex is at the origin
This equation is known as Euler's polyhedron formula. Thus the number of vertices is 2 more than the excess of the number of edges over the number of faces. For example, since a cube has 12 edges and 6 faces, the formula implies that it has eight vertices.
The four roots of the depressed quartic x 4 + px 2 + qx + r = 0 may also be expressed as the x coordinates of the intersections of the two quadratic equations y 2 + py + qx + r = 0 and y − x 2 = 0 i.e., using the substitution y = x 2 that two quadratics intersect in four points is an instance of Bézout's theorem.
Homogeneous coordinates are not uniquely determined by a point, so a function defined on the coordinates, say (,,), does not determine a function defined on points as with Cartesian coordinates. But a condition f ( x , y , z ) = 0 {\displaystyle f(x,y,z)=0} defined on the coordinates, as might be used to describe a curve, determines a condition ...
Translate the axes so that the vertex of the catenary lies on the y-axis and its height a is adjusted so the catenary satisfies the standard equation of the curve = and let the coordinates of P 1 and P 2 be (x 1, y 1) and (x 2, y 2) respectively. The curve passes through these points, so the difference of height is
Its x coordinate is half that of D, that is, x/2. The slope of the line BE is the quotient of the lengths of ED and BD, which is x 2 / x/2 = 2x. But 2x is also the slope (first derivative) of the parabola at E. Therefore, the line BE is the tangent to the parabola at E.