Over what interval is the function increasing, and over what interval is the function decreasing? x & f(x)=-(4)/(7) x^2 & (x, y) -14 & -112 & (-14,-112) -7 & -28 & (-7,-28) 0 & 0 & (0,0) 7 & -28 & (7,-28) 14 & -112 & (14,-112) The function f(x) is increasing over the interval square Simplify your answer. Type an inequality.)

The function \( f(x) = -\frac{4}{7} x^2 \) is a quadratic function with a negative leading coefficient, which means it is a downward-opening parabola. The vertex of this parabola is at the origin (0, 0), and the function decreases on either side of the vertex.<br /><br />For a downward-opening parabola, the function is increasing on the interval to the left of the vertex and decreasing on the interval to the right of the vertex. Therefore, the function is increasing over the interval:<br /><br />\[ (-\infty, 0) \]<br /><br />And it is decreasing over the interval:<br /><br />\[ (0, \infty) \]