Question

over what interval is the function increasing, and over what interval is the function decreasing?
the function ( f(x) ) is increasing over the interval (\boxed{}) (simplify your answer. type an inequality.)

Explanation:

Step1: Analyze the function type

The function \( f(x) = -8x^2 \) is a quadratic function. The general form of a quadratic function is \( f(x)=ax^2 + bx + c \), where \( a=-8 \), \( b = 0 \), \( c = 0 \). Since \( a=-8<0 \), the parabola opens downward. The vertex of the parabola \( f(x)=ax^2+bx + c \) is at \( x=-\frac{b}{2a} \). For \( f(x)=-8x^2 \), \( b = 0 \), so the vertex is at \( x = 0 \).

Step2: Determine increasing interval

For a parabola that opens downward ( \( a<0 \) ), the function is increasing to the left of the vertex (where \( x \) is less than the \( x \)-coordinate of the vertex) and decreasing to the right of the vertex (where \( x \) is greater than the \( x \)-coordinate of the vertex). The vertex is at \( x = 0 \), so the function is increasing when \( x<0 \).

Answer:

\( x < 0 \) (or in interval notation, \( (-\infty, 0) \), but since the question asks for an inequality, \( x < 0 \))