Question

Refer to the zeros of j(x) = 12x^3/5 - 27x/5 that you found in Question 2 of
Finding the Zeros of a Polynomial Function. Choose x-values and determine
what the graph of j(x) will look like between its zeros. In particular, make sure
to determine whether the graph will be above the x-axis or below the x-axis in
each region

Answer 1

To analyze the function \( j(x) = \frac{12x^3}{5} - \frac{27x}{5} \) and determine the behavior of its graph between its zeros, we first need to find the zeros of the function. We can set the function equal to zero:

\[
\frac{12x^3}{5} - \frac{27x}{5} = 0
\]

Multiplying both sides by 5 to eliminate the fraction:

\[
12x^3 - 27x = 0
\]

Factoring out a common term:

\[
3x(4x^2 - 9) = 0
\]

This gives us:

1. \( 3x = 0 \) which implies \( x = 0 \)
2. \( 4x^2 - 9 = 0 \) can be factored as \( (2x - 3)(2x + 3) = 0 \), giving \( x = \frac{3}{2} \) and \( x = -\frac{3}{2} \)

Thus, the zeros of \( j(x) \) are \( x = -\frac{3}{2}, 0, \frac{3}{2} \).

### Intervals to Test

The zeros divide the x-axis into the following intervals:

1. \( (-\infty, -\frac{3}{2}) \)
2. \( (-\frac{3}{2}, 0) \)
3. \( (0, \frac{3}{2}) \)
4. \( (\frac{3}{2}, \infty) \)

We can choose test points from each interval to determine whether \( j(x) \) is above or below the x-axis.

### Choose Test Points

1. **For the interval \( (-\infty, -\frac{3}{2}) \):**
- Choose \( x = -2 \)
- Calculate \( j(-2) = \frac{12(-2)^3}{5} - \frac{27(-2)}{5} = \frac{12(-8)}{5} + \frac{54}{5} = \frac{-96 + 54}{5} = \frac{-42}{5} < 0 \)
- Conclusion: The graph is **below** the x-axis.

2. **For the interval \( (-\frac{3}{2}, 0) \):**
- Choose \( x = -1 \)
- Calculate \( j(-1) = \frac{12(-1)^3}{5} - \frac{27(-1)}{5} = \frac{12(-1)}{5} + \frac{27}{5} = \frac{-12 + 27}{5} = \frac{15}{5} = 3 > 0 \)
- Conclusion: The graph is **above** the x-axis.

3. **For the interval \( (0, \frac{3}{2}) \):**
- Choose \( x = 1 \)
- Calculate \( j(1) = \frac{12(1)^3}{5} - \frac{27(1)}{5} = \frac{12}{5} - \frac{27}{5} = \frac{-15}{5} = -3 < 0 \)
- Conclusion: The graph is **below** the x-axis.

4. **For the interval \( (\frac{3}{2}, \infty) \):**
- Choose \( x = 2 \)
- Calculate \( j(2) = \frac{12(2)^3}{5} - \frac{27(2)}{5} = \frac{12(8)}{5} - \frac{54}{5} = \frac{96 - 54}{5} = \frac{42}{5} > 0 \)
- Conclusion: The graph is **above** the x-axis.

### Summary of the Graph Behavior Between Zeros

- In the interval \( (-\infty, -\frac{3}{2}) \): **Below the x-axis**
- In the interval \( (-\frac{3}{2}, 0) \): **Above the x-axis**
- In the interval \( (0, \frac{3}{2}) \): **Below the x-axis**
- In the interval \( (\frac{3}{2}, \infty) \): **Above the x-axis**

This analysis shows how the function transitions across its zeros and how it behaves in each interval.