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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.
For example, the graph of y = x 2 − 4x + 7 can be obtained from the graph of y = x 2 by translating +2 units along the X axis and +3 units along Y axis. This is because the equation can also be written as y − 3 = (x − 2) 2. For many trigonometric functions, the parent function is usually a basic sin(x), cos(x), or tan(x).
In the graph, moving one unit to the right (increasing x by 1) moves the y-value up by a: that is, (+) = +. Negative slope a indicates a decrease in y for each increase in x . For example, the linear function y = − 2 x + 4 {\displaystyle y=-2x+4} has slope a = − 2 {\displaystyle a=-2} , y -intercept point ( 0 , b ) = ( 0 , 4 ...
Given two different points (x 1, y 1) and (x 2, y 2), there is exactly one line that passes through them. There are several ways to write a linear equation of this line. If x 1 ≠ x 2, the slope of the line is . Thus, a point-slope form is [3]
The solution set for the equations x − y = −1 and 3x + y = 9 is the single point (2, 3). A solution of a linear system is an assignment of values to the variables ,, …, such that each of the equations is satisfied. The set of all possible solutions is called the solution set. [5]
A cycle graph or circular graph of order n ≥ 3 is a graph in which the vertices can be listed in an order v 1, v 2, …, v n such that the edges are the {v i, v i+1} where i = 1, 2, …, n − 1, plus the edge {v n, v 1}. Cycle graphs can be characterized as connected graphs in which the degree of all vertices is 2.
A graph is planar if it contains as a subdivision neither the complete bipartite graph K 3,3 nor the complete graph K 5. Another problem in subdivision containment is the Kelmans–Seymour conjecture: Every 5-vertex-connected graph that is not planar contains a subdivision of the 5-vertex complete graph K 5.
Consider the case where Y is the graph with vertex set {1,2,3} and undirected edges {1,2}, {1,3} and {2,3} (a triangle or 3-circle) with vertex states from K = {0,1}. For vertex functions use the symmetric, boolean function nor : K 3 → K defined by nor(x,y,z) = (1+x)(1+y)(1+z) with boolean arithmetic. Thus, the only case in which the function ...