Question

Refer to the polynomial function h(x)=-3x^2 (5x+10)(x-2) that you wrote in question 1 of Finding the Zeros of a Polynomial Function. Describe and provide an explanation for the end behavior of h(x).

Answer 1

To analyze the end behavior of the polynomial function \( h(x) = -3x^2(5x + 10)(x - 2) \), we first examine the highest degree term of the polynomial, as it primarily determines the end behavior as \( x \) approaches positive and negative infinity.

1. **Identify the Leading Term:**
- The polynomial \( h(x) = -3x^2(5x + 10)(x - 2) \) can be expanded to find the leading term. However, we can analyze it without fully expanding by understanding the components.
- The term \( x^2 \) suggests that our function is a degree 4 polynomial because \( 5x + 10 \) contributes a degree of 1 and \( x - 2 \) contributes a degree of 1, resulting in \( 2 + 1 + 1 = 4 \).

2. **Leading Coefficient:**
- The leading coefficient will come from the coefficients of \( -3x^2(5x)(x) \) when fully expanded:
- The leading term from \( -3x^2 \) is \( -3x^2 \).
- The term \( (5x + 10) \) and \( (x - 2) \) will contribute \( 5x^1 \) and \( x^1 \), respectively.
- Thus, the leading term when fully multiplied out will be \( -3 \cdot 5 \cdot 1 \cdot x^4 = -15x^4 \).

3. **End Behavior:**
- Since the leading term is \( -15x^4 \):
- As \( x \to +\infty \) (as x approaches positive infinity), \( h(x) \to -\infty \) because raising a positive number to the power of 4 yields a positive result, which is then multiplied by the negative leading coefficient.
- As \( x \to -\infty \) (as x approaches negative infinity), \( h(x) \to -\infty \) again because raising a negative number also to the power of 4 yields a positive result, multiplied by the negative leading coefficient still causes the output to go to negative infinity.

4. **Conclusion:**
- The end behavior of \( h(x) \) is that \( h(x) \) approaches \( -\infty \) as \( x \to +\infty \) and also approaches \( -\infty \) as \( x \to -\infty \).
- In summary, for large absolute values of \( x \) (in either direction), the polynomial \( h(x) \) continues to decrease without bound.