For the function \( f(x) \), find the maximum number of real zeros, the maximum number of \( x \)-intercepts, and the maximum number of turning points that the function can have. \( f(x)=-6 x-x^{3} \) \( f(x) \) has a maximum of 3 real zeros. \( f(x) \) has a maximum of 0 tuming points.

To dive into the world of polynomials, let's examine the given function \( f(x) = -6x - x^3 \). When we analyze this function, we see it's a cubic polynomial. Cubic polynomials can have up to three real zeros, corresponding to the points where the graph intersects the x-axis. So, yes, you are absolutely right that \( f(x) \) can have a maximum of 3 real zeros! Regarding the x-intercepts, these are simply the same as the real zeros. Thus, the maximum number of x-intercepts for \( f(x) \) is also 3. As for the turning points, a cubic polynomial can have a maximum of 2 turning points where the function changes direction. However, since the leading coefficient is negative, the function will decrease to the right and may not turn back up, depending on the critical points you compute. So, generally, expect a maximum of 2 turning points! In summary: - **Maximum real zeros**: 3 - **Maximum x-intercepts**: 3 - **Maximum turning points**: 2 (opposite to your note of 0, which could have been a mix-up). Keep swirling through the algebraic world!