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Find the domain and range of f(x)=x^2-5x+6
Bot
answered
The function is a quadratic function, which means that the domain is all real numbers, since the function exists for any value of x.
To find the range, we can use the fact that the vertex of the parabola is at x = -b/2a = 5/2.
Since the coefficient of x^2 is positive, the parabola opens upwards, and the minimum value occurs at the vertex.
Substituting x = 5/2 into the original equation, we get f(5/2) = (5/2)^2 - 5(5/2) + 6 = -1/4.
Therefore, the range of the function is (-1/4, infinity).
Step-by-Step Bot
answered
To find the domain and range of the function f(x) = x^2 - 5x + 6, we will analyze its characteristics.
Domain:
The domain of a function is the set of all possible input values (x-values) for which the function is defined. In this case, f(x) is a polynomial function, and polynomials are defined for all real numbers. Hence, the domain of f(x) is all real numbers, or (-∞, ∞).
Range:
The range of a function is the set of all possible output values (y-values). To determine the range of f(x), we need to find the minimum/maximum value of the function.
First, let's consider the coefficient of the x^2 term, which is positive. This indicates that the parabola opens upwards, meaning there is no maximum value. Additionally, since there is no x-term coefficient, the function does not shift vertically.
Next, we look for the x-coordinate of the vertex. Using the formula x = -b/2a (where a is the coefficient of x^2, and b is the coefficient of x), we find x = -(-5) / (2 * 1) = 5/2.
To find the corresponding y-coordinate (the value of f(x) at the vertex), substitute this x-value back into the function: f(5/2) = (5/2)^2 - 5(5/2) + 6 = 1/4.
Therefore, the vertex of the parabola is (5/2, 1/4). Since the parabola opens upwards, the minimum value occurs at the vertex.
Hence, the range of f(x) is [1/4, ∞).