Question

over what interval is the function increasing, and over what interval is the function decreasing?
x | f(x)=0.3x² | (x,y)
-2 | 1.2 | (-2,1.2)
-1 | 0.3 | (-1,0.3)
0 | 0 | (0,0)
1 | 0.3 | (1,0.3)
2 | 1.2 | (2,1.2)
the function f(x) is increasing over the interval x>0.
(simplify your answer. type an inequality.)
the function f(x) is decreasing over the interval \square.
(simplify your answer. type an inequality.)

Explanation:

Step1: Analyze the function's behavior

The function is \( f(x) = 0.3x^2 \), which is a quadratic function with a positive leading coefficient, so it opens upwards. The vertex is at \( x = 0 \) (since for \( ax^2+bx+c \), the vertex is at \( x = -\frac{b}{2a} \), here \( b = 0 \), so \( x = 0 \)).

Step2: Determine decreasing interval

For a parabola opening upwards, the function decreases to the left of the vertex ( \( x < 0 \)) and increases to the right of the vertex ( \( x > 0 \)). Looking at the table: when \( x \) goes from -2 to 0 (values like -2, -1, 0), the \( f(x) \) values decrease from 1.2 to 0.3 to 0. So the function is decreasing when \( x < 0 \).

Answer:

The function \( f(x) \) is decreasing over the interval \( x < 0 \).