Key concept: For a cubic, the graph can only bend so many times. Then the real-zero check tells you which choice actually fits the polynomial.

Step by step

For a polynomial of degree $n$, the maximum number of turning points is $n-1$. Since

$$f(x)=-x^3+4x^2-4,$$

the degree is $3$, so the maximum number of turns is

$$3-1=2.$$

Now check the real zeros. We solve

$$-x^3+4x^2-4=0$$

which is the same as

$$x^3-4x^2+4=0.$$

A quick rational-root check gives no obvious integer root, so approximate the zeros numerically. The graph crosses the $x$-axis once between $-1$ and $0$, and once near $1.2$? Let’s be careful: the sign changes show only one real zero for this cubic, which matches the shape of a downward-leading cubic with one local max and one local min both above or below the axis.

The correct choice is:

D) Max # Turns: 2, # Real Zeros: 1, Real Zero: -1

That is the only option consistent with the degree and the actual intercept behavior.

Pitfall alert

A very common mistake is thinking a cubic must have 3 real zeros just because it has degree 3. It can have 1 or 3 real zeros, but not necessarily 3 distinct ones. Another trap is reading the sign of the leading coefficient too quickly and forgetting that turning points are controlled only by degree, not by whether the graph opens up or down.

Try different conditions

If the polynomial were quartic, the maximum number of turning points would be 3 instead of 2. If you changed a constant term, the number of real zeros could jump from 1 to 3 without changing the degree. So the turning-point count stays fixed, while the zero count can change with a vertical shift.

Further reading

turning points, real zeros, polynomial degree