Search results
Results from the WOW.Com Content Network
Given a function: from a set X (the domain) to a set Y (the codomain), the graph of the function is the set [4] = {(, ()):}, which is a subset of the Cartesian product.In the definition of a function in terms of set theory, it is common to identify a function with its graph, although, formally, a function is formed by the triple consisting of its domain, its codomain and its graph.
The reciprocal function: y = 1/x.For every x except 0, y represents its multiplicative inverse. The graph forms a rectangular hyperbola.. In mathematics, a multiplicative inverse or reciprocal for a number x, denoted by 1/x or x −1, is a number which when multiplied by x yields the multiplicative identity, 1.
Graphs of y = b x for various bases b: base 10, base e, base 2, base 1 / 2 . Each curve passes through the point (0, 1) because any nonzero number raised to the power of 0 is 1. At x = 1, the value of y equals the base because any number raised to the power of 1 is the number itself.
The roots of the quadratic function y = 1 / 2 x 2 − 3x + 5 / 2 are the places where the graph intersects the x-axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
The graph of the logarithm base 2 crosses the x-axis at x = 1 and passes through the points (2, 1), (4, 2), and (8, 3), depicting, e.g., log 2 (8) = 3 and 2 3 = 8. The graph gets arbitrarily close to the y-axis, but does not meet it. Addition, multiplication, and exponentiation are three of the most fundamental arithmetic operations.
[1] [4] [5] For example, the equations = = form a parametric representation of the unit circle, where t is the parameter: A point (x, y) is on the unit circle if and only if there is a value of t such that these two equations
The cube of every connected graph necessarily contains a Hamiltonian cycle. [10] It is not necessarily the case that the square of a connected graph is Hamiltonian, and it is NP-complete to determine whether the square is Hamiltonian. [11] Nevertheless, by Fleischner's theorem, the square of a 2-vertex-connected graph is always Hamiltonian. [12]
At any red point, the problem is polynomial-time computable; at any blue point, the problem is #P-hard in general, but polynomial-time computable for planar graphs; and at any point in the white regions, the problem is #P-hard even for bipartite planar graphs. If both x and y are non-negative integers, the problem (,) belongs to #P.