To analyze the graph of the function \( h(x) = -3x^2(5x + 5)(x - 1) \), we first need to determine its zeros and how it behaves in the intervals defined by these zeros.
Step 1: Find the Zeros
The zeros of the function \( h(x) = -3x^2(5x + 5)(x - 1) \) can be found by setting the function equal to zero:
\[ -3x^2(5x + 5)(x - 1) = 0 \]
This gives us three factors to consider:
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\( -3x^2 = 0 \)
- This yields \( x = 0 \).
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\( 5x + 5 = 0 \)
- Solving this gives: \[ 5x = -5 \implies x = -1 \]
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\( x - 1 = 0 \)
- This gives \( x = 1 \).
Thus, the zeros of the function are \( x = -1, x = 0, \) and \( x = 1 \).
Step 2: Determine Intervals
Next, we analyze the intervals defined by these zeros:
- \( (-\infty, -1) \)
- \( (-1, 0) \)
- \( (0, 1) \)
- \( (1, \infty) \)
Step 3: Test Points in Each Interval
We will pick a test point in each interval to determine whether \( h(x) \) is positive or negative:
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Interval \( (-\infty, -1) \): Choose \( x = -2 \) \[ h(-2) = -3(-2)^2(5(-2) + 5)(-2 - 1) = -3(4)(-10 + 5)(-3) = -3(4)(-5)(-3) = -180 \quad (\text{negative}) \]
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Interval \( (-1, 0) \): Choose \( x = -0.5 \) \[ h(-0.5) = -3(-0.5)^2(5(-0.5) + 5)(-0.5 - 1) = -3(0.25)(-2.5 + 5)(-1.5) = -3(0.25)(2.5)(-1.5) = 2.8125 \quad (\text{positive}) \]
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Interval \( (0, 1) \): Choose \( x = 0.5 \) \[ h(0.5) = -3(0.5)^2(5(0.5) + 5)(0.5 - 1) = -3(0.25)(2.5 + 5)(-0.5) = -3(0.25)(7.5)(-0.5) = 2.8125 \quad (\text{positive}) \]
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Interval \( (1, \infty) \): Choose \( x = 2 \) \[ h(2) = -3(2)^2(5(2) + 5)(2 - 1) = -3(4)(10 + 5)(1) = -3(4)(15) = -180 \quad (\text{negative}) \]
Summary of Results
Now we summarize the behavior of \( h(x) \):
- In the interval \( (-\infty, -1) \), \( h(x) < 0 \) (below the x-axis).
- In the interval \( (-1, 0) \), \( h(x) > 0 \) (above the x-axis).
- In the interval \( (0, 1) \), \( h(x) > 0 \) (above the x-axis).
- In the interval \( (1, \infty) \), \( h(x) < 0 \) (below the x-axis).
Graph Behavior Summary
The graph of \( h(x) \) crosses the x-axis at \( x = -1, 0, 1 \). It will be below the x-axis for \( x < -1 \) and \( x > 1 \), and above the x-axis for \( -1 < x < 0 \) and \( 0 < x < 1 \). The function \( h(x) \) has a structure showing local maxima and minima in the intervals defined based on these zeros.
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