Final answer:
The polynomial f(x)=8(x-5)(x+9)^2 has real zeros at x = 5 (multiplicity of 1) and x = -9 (multiplicity of 2). It crosses the x-axis at x = 5 and touches at x = -9. The maximum number of turning points is 2, and for large values, the graph resembles y = 8x^3.
Step-by-step explanation:
For the polynomial function f(x)=8(x−5)(x+9)^2, we can address each part of the question as follows:
- Real zeros and their multiplicities: This polynomial has two real zeros, x = 5 and x = -9. The zero at x = 5 has a multiplicity of 1 since it appears only once in the factorized form, and the zero at x = -9 has a multiplicity of 2 because it appears twice.
- Graph behavior at x-intercepts: The graph will cross the x-axis at x = 5 because the multiplicity is odd. At x = -9, the graph will touch but not cross the x-axis, as the zero has an even multiplicity.
- Maximum number of turning points: A polynomial of degree n can have up to n-1 turning points. Since f(x) is a cubic polynomial (degree 3), it can have at most 2 turning points.
- End behavior of f(x): The end behavior is similar to that of the leading term, which is 8x^3. As x approaches positive or negative infinity, the graph of f(x) will resemble that of y = 8x^3, moving towards positive infinity for both directions.