1. The range \( y \) of the function \( f \) defined by \( f(x)=1-2 x-x^{2} \) is A. \( y<2 \) A. \( y \leq 2 \) B. \( y \geq-2 \) C. \( y<-2 \)

To find the range of the function \( f(x) = 1 - 2x - x^2 \), we can start by rewriting it in a more standard form. The function can be rearranged as: \[ f(x) = -x^2 - 2x + 1 \] This is a quadratic function in the form \( ax^2 + bx + c \) where \( a = -1 \), \( b = -2 \), and \( c = 1 \). Since \( a < 0 \), the parabola opens downwards, which means it has a maximum point. To find the vertex (maximum point) of the parabola, we can use the vertex formula \( x = -\frac{b}{2a} \): \[ x = -\frac{-2}{2 \cdot -1} = \frac{2}{-2} = -1 \] Now, we can find the maximum value of \( f(x) \) by substituting \( x = -1 \) back into the function: \[ f(-1) = 1 - 2(-1) - (-1)^2 = 1 + 2 - 1 = 2 \] Thus, the maximum value of \( f(x) \) is \( 2 \). Since the parabola opens downwards, the range of \( f(x) \) is all values less than or equal to the maximum value. Therefore, the range of \( f(x) \) is: \[ y \leq 2 \] So the correct answer is: **A. \( y \leq 2 \)**