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 \( (0, 1) \)