Refer to the polynomial function h(x) = -3x^2(5x+3)(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 $h(x)=-3x^2(5x+3)(x-2)$, we first need to determine the degree and leading coefficient of the polynomial.

Step 1: Find the degree of the polynomial

The function can be expressed as the product of its factors. We have:

1. The term $-3x^2$, which has a degree of 2.
2. The term $(5x+3)$, which has a degree of 1.
3. The term $(x-2)$, which has a degree of 1.

To find the total degree of $h(x)$, we add the degrees of all terms:

$$\text{Degree}=2+1+1=4$$

Step 2: Identify the leading coefficient

The leading term of the polynomial comes from multiplying the leading coefficients of each factor. The leading term from $-3x^2$ is $-3x^2$, and the leading term from the polynomial expression $(5x+3)(x-2)$ is $5x^2$ (since $5x\cdot x$). Thus, the leading term of $h(x)$ is:

$$-3x^2\cdot 5x^2=-15x^4$$

So, the leading coefficient is $-15$.

Step 3: Determine end behavior based on degree and leading coefficient

Now that we have established that $h(x)$ is a polynomial of degree 4 (even degree) with a leading coefficient of $-15$ (negative), we can analyze the end behavior:

• For polynomials with an even degree:
  • As $x$ approaches $+\mathrm{\infty }$ (right end), $h(x)$ approaches $-\mathrm{\infty }$ (downward).
  • As $x$ approaches $-\mathrm{\infty }$ (left end), $h(x)$ also approaches $-\mathrm{\infty }$ (downward).

Conclusion

The end behavior of the polynomial function $h(x)=-3x^2(5x+3)(x-2)$ is described as follows:

• As $x\to +\mathrm{\infty },h(x)\to -\mathrm{\infty }$
• As $x\to -\mathrm{\infty },h(x)\to -\mathrm{\infty }$

This means that both ends of the graph of $h(x)$ will trend downwards as you move to the left or right.