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y = -4x2 + 1

Range = {y:y=<1, y is an element of the reals}

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Q: What is the range of the function y -4x2 1?
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Related questions

What are function and relation?

A function is an equation that gives a unique answer. A relation does not. Example: y = 3x + 1 is a function. If I give you x, you can determine y. And that y is unique to that x. So if x = 1, you know y = 4. No other of x gives y = 4 as an answer. So y = 3x + 1 is a function. Example: y = 4x2. So if I give you x = 1, y = 4. But y = 4 if I also give you x = -1. So y = 4x2 is not a function, it is a relation.


How is the graph of y 4x2 1 different from the graph of y 4x2?

Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Assuming the first function is y = 4x2 + 1, and the second y = 4x2 then the first graph is 1 unit higher than the second.


What is the range of the equation y 4x2 - 8?

y &gt; -8


How do you write 4x2 plus y equals 16 using function notation?

First you need to solve for y. So write 4x2+y=16 so y=16-4x2 Now write f(x)=16-4x2


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

y &lt; 1


What is the domain of the continuous quadratic function y equals 4x2 plus 2?

The domain is from negative infinity to positive infinity. The range is from positive 2 to positive infinity.


What is the range of the secant function?

absolute value of y&gt; 1


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

y = 2sin(x)? If that's your function, well we know that sin(x) oscillates between y = 1 and y = -1, but in our case we have double that from 2sin(x), so our range is -2 to 2.


Is y equals 4x2-1 a linear equation?

yes


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.


Is 4 x squared opens upward?

We have f(x) = y = 4x2 When x = 1, f(1) = 4(1)2 = 4 and when x = -1, f(-1) = 4(-1)2 = 4 So for negative and positive values of x, y is always positive. So y = 4x2 opens upward. Visit the link mentioned below to know more about the given function.


How do you factor 4x2 - y2?

4x2-y2 = (2x-y)(2x+y)