Question

Choose x- values and determine what the graph of h(x) = -3x^2(5x + 5)(x-1) 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. Show your work.

Answer 1

To analyze the behavior of the function \( h(x) = -3x^2(5x + 5)(x-1) \) between its zeros, we first need to find the zeros of the function by setting \( h(x) = 0 \).

### Step 1: Finding the Zeros

1. **Factor \( h(x) \):**

We can factor \( -3x^2 \) out of the function, as well as \( 5(x + 1) \) from \( 5x + 5 \):
\[
h(x) = -3x^2(5(x + 1))(x - 1)
\]

2. **Set each factor to zero:**

- From \( -3x^2 = 0 \):
\[
x = 0
\]
- From \( 5(x + 1) = 0 \):
\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\]
- From \( (x - 1) = 0 \):
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]

The zeros of the function \( h(x) \) are \( x = -1, 0, 1 \).

### Step 2: Determine Intervals for Sign Analysis

The zeros divide the x-axis into the following intervals:
- \( (-\infty, -1) \)
- \( (-1, 0) \)
- \( (0, 1) \)
- \( (1, \infty) \)

### Step 3: Test Points in Each Interval

We will choose test points in each of these intervals to determine the sign of \( h(x) \).

1. **Interval \( (-\infty, -1) \)**: Choose \( x = -2 \)
\[
h(-2) = -3(-2)^2(5(-2 + 1))(-2 - 1) = -3(4)(-5)(-3)
\]
Since there are three negative factors, the product is negative:
\[
h(-2) < 0 \quad \text{(below the x-axis)}
\]

2. **Interval \( (-1, 0) \)**: Choose \( x = -0.5 \)
\[
h(-0.5) = -3(-0.5)^2(5(-0.5 + 1))(-0.5 - 1) = -3(0.25)(2.5)(-1.5)
\]
This product has two negative factors and one positive factor, so the overall product is positive:
\[
h(-0.5) > 0 \quad \text{(above the x-axis)}
\]

3. **Interval \( (0, 1) \)**: Choose \( x = 0.5 \)
\[
h(0.5) = -3(0.5)^2(5(0.5 + 1))(0.5 - 1) = -3(0.25)(7.5)(-0.5)
\]
This has two negative factors and one positive factor, so the overall product is positive:
\[
h(0.5) > 0 \quad \text{(above the x-axis)}
\]

4. **Interval \( (1, \infty) \)**: Choose \( x = 2 \)
\[
h(2) = -3(2)^2(5(2 + 1))(2 - 1) = -3(4)(15)(1)
\]
Since there is only one negative factor (the \( -3 \)), the overall product is negative:
\[
h(2) < 0 \quad \text{(below the x-axis)}
\]

### Summary of the Sign in Each Region:

- For \( x \in (-\infty, -1) \), \( h(x) < 0 \) (below the x-axis)
- For \( x \in (-1, 0) \), \( h(x) > 0 \) (above the x-axis)
- For \( x \in (0, 1) \), \( h(x) > 0 \) (above the x-axis)
- For \( x \in (1, \infty) \), \( h(x) < 0 \) (below the x-axis)

### Conclusion:

The function \( h(x) \) will be:
- Below the x-axis on \( (-\infty, -1) \) and \( (1, \infty) \)
- Above the x-axis on \( (-1, 0) \) and \( (0, 1) \)