Question

Identify the vertex: The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by the x-coordinate x_v = -b/(2a). In this case, a = -1, b = 4, so x_v = -4/(2 * (-1)) = 2. To find the y-coordinate (the maximum value), plug this x-coordinate back into the function: f(2) = -2^2 + 4*2 - 3 = -4 + 8 - 3 = 1. So, the vertex is at (2, 1).

The maximum value of the function occurs at the vertex, which is (2, 1) in this case. Therefore, the maximum value of f(x) is 1.

Answer 1

The vertex of a quadratic function can be found using the formula x_v = -b/(2a). In this case, the vertex is at (2, 1) with a maximum value of 1.

Step-by-step explanation:

The vertex of a quadratic function in the form f(x) = ax² + bx + c can be found using the formula x_v = -b/(2a), where a, b, and c are the coefficients of the function. In this case, a = -1, b = 4, and c = -3. Plugging these values into the formula, we get x_v = -4/(2*(-1)) = 2. To find the y-coordinate of the vertex, we substitute the x-coordinate back into the function: f(2) = (-2)^2 + 4*2 - 3 = 1. Therefore, the vertex of the function is at (2, 1), and the maximum value of f(x) is 1.