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Q: What is the range of the function y x 2?
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What is the range of the function y 2sin x?

The answer will range between '2' & '-2' Reason; The Sine function ranges between '1' & '-1' , so if it has a coefficient of '2', this will increase the range to '2' & '-2'.


What is the domain and range of the sine function y is equal to 2 sin x?

Domain (input or 'x' values): -&infin; < x < &infin;.Range (output or 'y' values): -2 &le; y &le; 2.


What is the range of the function y equals -x 2 plus 1?

y &lt; 1


What is the range of the function y equals x?

The function y=x is a straight line. The range is all real numbers.


Domain is 2 range is 2 is this a function?

yes y=x Like 2=2


What is the range of the following quadratic function y 3(x -2)2 plus 5?

It is y &gt;= 5.


What is the range of the function yx2?

The range of the function y=x^2 would be y is greater than or equal to 0 in this case. So pretty much just find the vertex of the function and what ever the y coordinate is set that as the lowest number for the range.


In the ordered pair x y the value of y is a member of the?

x is a member of the function's domain, y is a member of the function's range.


What is the range of a linear function?

The range is the y, while the domain is the x.


Does the graph x-y2 equals 1 represent x as a function of y?

X - Y^2 = 1 - Y^2 = - X + 1 Y^2 = X - 1 Y = (+/-) sqrt(X - 1) now, X is represented as a function of Y. Function values are generally Y values.


What is the range of the function y x?

The function y=x is a straight line. The range is all real numbers.The functions just tend to infinity as the x and y values get infinitely large or infinitely small.


If an inverse function undoes the work of the original function the original functions range becomes the inverse functions?

Maybe; the range of the original function is given, correct? If so, then calculate the range of the inverse function by using the original functions range in the original function. Those calculated extreme values are the range of the inverse function. Suppose: f(x) = x^3, with range of -3 to +3. f(-3) = -27 f(3) = 27. Let the inverse function of f(x) = g(y); therefore g(y) = y^(1/3). The range of f(y) is -27 to 27. If true, then f(x) = f(g(y)) = f(y^(1/3)) = (y^(1/3))^3 = y g(y) = g(f(x)) = g(x^3) = (x^3)^3 = x Try by substituting the ranges into the equations, if the proofs hold, then the answer is true for the function and the range that you are testing. Sometimes, however, it can be false. Look at a transcendental function.