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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.
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.
If b = 0, the line is a vertical line (that is a line parallel to the y-axis) of equation =, which is not the graph of a function of x. Similarly, if a ≠ 0, the line is the graph of a function of y, and, if a = 0, one has a horizontal line of equation =.
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 ...
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]
The unit circle can be specified as the level curve f(x, y) = 1 of the function f(x, y) = x 2 + y 2.Around point A, y can be expressed as a function y(x).In this example this function can be written explicitly as () =; in many cases no such explicit expression exists, but one can still refer to the implicit function y(x).
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If y 2 = x 3 − x − 1, then the field C(x, y) is an elliptic function field. The element x is not uniquely determined; the field can also be regarded, for instance, as an extension of C(y). The algebraic curve corresponding to the function field is simply the set of points (x, y) in C 2 satisfying y 2 = x 3 − x − 1.