Negative values can be used for x, but the range is restricted because x 2 ≥ 0. The correct answer is: The domain is all real numbers and the range is all real numbers f( x) such that f( x) ≥ 7. Quadratic functions have no domain restrictions. There is no value of x for which, so this proves that the range is restricted.įind the domain and range of the real-valued function f( x) = x 2 + 7.Ī) The domain is all real numbers and the range is all real numbers f( x) such thatī) The domain is all real numbers x such that x ≥ 0 and the range is all real numbers f(x) such that f( x) ≥ 7.Ĭ) The domain is all real numbers x such that x ≥ 0 and the range is all real numbers.ĭ) The domain and range are all real numbers.Ĭorrect. Since the attempt to solve ends with a false statement-0 cannot be equal to 6!-the equation has no solution. Is it possible for to be equal to 3? Write an equation and try to solve it. You can check the horizontal asymptote, y = 3. The domain is all real numbers except −2, and the range is all real numbers except 3. The other, a horizontal asymptote, appears to be around y = 3. One asymptote, a vertical asymptote, is at x =−2, as you should expect from the domain restriction. In this case, x + 2 is the denominator, and this is 0 only when x = −2.įor the range, create a graph using a graphing utility and look for asymptotes: The domain of a rational function is restricted where the denominator is 0. What are the domain and range of the real-valued function ? Evaluating f(1) gives f(1) = 4, so the vertex is at (1, 4). The line of reflection here is x = 1, so the vertex must be at the point (1, f(1)). (You can verify this by evaluating f(2) and f(0).) That is, both (2, 1) and (0, 1) are on the graph. In the previous example, notice that when x = 2 and when x = 0, the function value is 1. The vertex must lie on the line of reflection, because it’s the only point that does not have a mirror image! Since a quadratic function has two mirror image halves, the line of reflection has to be in the middle of two points with the same y value. You can check that the vertex is indeed at (1, 4). The domain is all real numbers, and the range is all real numbers f( x) such that f( x) ≤ 4. From the graph, you can see that f( x) ≤ 4. The vertex, or turning point, is at (1, 4). With quadratic functions, remember that there is either a maximum (greatest) value, or a minimum (least) value. Any real number can be used for x to get a meaningful output.īecause the coefficient of x 2 is negative, it will open downward. There are no rational or radical expressions, so there is nothing that will restrict the domain. What are the domain and range of the real-valued function f( x) = −3 x 2 + 6 x + 1?
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