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. \( f(x)=-6 x-x^{3} \)

We start with the given function: \[ f(x)=-6x-x^3. \] **Step 1. Factor the Function** Factor out the common term: \[ f(x)=-x(x^2+6). \] **Step 2. Find the Real Zeros** Set \( f(x)=0 \): \[ -x(x^2+6)=0. \] This equation is satisfied when either factor is zero: - For \(-x=0\), we get \( x=0 \). - For \(x^2+6=0\), we have: \[ x^2=-6. \] Since \(-6\) is negative, this equation has no real solutions. Thus, the only real zero is at \( x=0 \). **Step 3. Determine the Maximum Number of \(x\)-Intercepts** The \(x\)-intercepts are the points on the graph where \( f(x)=0 \). Since we found only one real zero, the graph has exactly one \(x\)-intercept at: \[ (0,0). \] **Step 4. Find the Turning Points** Turning points occur where the derivative of \( f(x) \) is zero. First, compute the derivative: \[ f'(x) = \frac{d}{dx}(-6x) + \frac{d}{dx}(-x^3) = -6 - 3x^2. \] Set the derivative equal to zero: \[ -6 - 3x^2=0. \] Solve for \( x \): \[ -3x^2=6 \quad\Longrightarrow\quad x^2=-2. \] Since there is no real solution to \( x^2=-2 \), the function has no turning points. **Final Answers:** - Maximum number of real zeros: \(1\) - Maximum number of \(x\)-intercepts: \(1\) - Maximum number of turning points: \(0\)