To analyze the end behavior of the polynomial function \( h(x) = -3x^2(5x + 5)(x - 1) \), we need to consider both its degree and the leading coefficient.
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Identify the degree:
- The term \( -3x^2 \) has a degree of 2.
- The term \( (5x + 5) \) is linear (degree 1).
- The term \( (x - 1) \) is also linear (degree 1).
- Therefore, the total degree of \( h(x) \) is: \[ 2 + 1 + 1 = 4 \]
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Identify the leading coefficient:
- The leading term of \( h(x) \) can be determined by multiplying the coefficients of the highest degree terms:
- The leading term from \( -3x^2 \) is \( -3x^2 \).
- The leading term from \( 5x + 5 \) is \( 5x \).
- The leading term from \( x - 1 \) is \( x \).
- When we multiply these together considering the leading degree: \[ \text{Leading term} = -3 \cdot 5 \cdot 1 \cdot x^2 \cdot x \cdot x = -15x^4 \]
- Thus, the leading coefficient of the polynomial is \( -15 \) and the leading degree is \( 4 \).
- The leading term of \( h(x) \) can be determined by multiplying the coefficients of the highest degree terms:
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Analyze the end behavior:
- For a polynomial of even degree (which \( 4 \) is) and a negative leading coefficient (which \( -15 \) is), the end behavior can be described as follows:
- As \( x \to \infty \) (x goes to positive infinity), \( h(x) \to -\infty \) (the function goes down).
- As \( x \to -\infty \) (x goes to negative infinity), \( h(x) \to -\infty \) (the function also goes down).
- For a polynomial of even degree (which \( 4 \) is) and a negative leading coefficient (which \( -15 \) is), the end behavior can be described as follows:
Therefore, the end behavior of the polynomial function \( h(x) = -3x^2(5x + 5)(x - 1) \) is that both ends of the graph (as \( x \) approaches \( \infty \) and \( -\infty \)) go down towards negative infinity, resulting in a graph that opens downwards.
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