Domain and range of f(x)-1/4x^2+2x+5
To find the domain and range of the function f(x) = -1/4x^2 + 2x + 5, we need to consider two things:
1. Domain: The domain of a function represents all possible values that x can take. In this case, since f(x) is a quadratic function, it is defined for all real numbers. Therefore, the domain of f(x) is (-∞, ∞).
2. Range: The range of a function represents all possible values that f(x) can take. To determine the range, we can consider the vertex of the quadratic function. The vertex is given by the formula x = -b/2a, where a, b, and c correspond to the coefficients of the quadratic equation in the form ax^2 + bx + c.
In this case, the quadratic function -1/4x^2 + 2x + 5 is in the form ax^2 + bx + c, where a = -1/4, b = 2, and c = 5. Plugging these values into the vertex formula, we get:
x = -b/2a = -2/(2*(-1/4)) = -2/(-1/2) = -2 * (-2) = 4
To find the corresponding y-coordinate of the vertex, we substitute x = 4 into the function:
f(4) = -1/4(4)^2 + 2(4) + 5 = -1/4(16) + 8 + 5 = -4 + 8 + 5 = 9
Therefore, the vertex of the function is (4, 9). Since this is a downward-opening parabola (a < 0), the range of the function is (-∞, 9].
So, the domain of f(x) is (-∞, ∞) and the range is (-∞, 9].