A polynomial function has a root of -7 with multiplicity 2, a root of -1 with multiplicity 1, a root of 2 with multiplicity 4, and a root of 4 with multiplicity 1. If the function has a positive leading coefficient and is of even degree, which statement about the graph is true?
A. The graph of the function is positive on $(2,4)$.
B. The graph of the function is negative on $(4, \infty)$.
C. The graph of the function is positive on $(-\infty,-7)$.
D. The graph of the function is negative on $(-7,-1)$.
Answer (2)
The polynomial function is of degree 8 with a positive leading coefficient. The graph is positive on ( − ∞ , − 7 ) and negative on ( − 7 , − 1 ) . The correct statement from the options is that the graph is positive on ( − ∞ , − 7 ) .
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Determine the degree of the polynomial by summing the multiplicities of the roots: 2 + 1 + 4 + 1 = 8 .
Write the general form of the polynomial as f ( x ) = a ( x + 7 ) 2 ( x + 1 ) ( x − 2 ) 4 ( x − 4 ) , where 0"> a > 0 is the leading coefficient.
Analyze the sign of f ( x ) on the intervals ( 2 , 4 ) , ( 4 , ∞ ) , ( − ∞ , − 7 ) , and ( − 7 , − 1 ) .
The graph of the function is positive on ( − ∞ , − 7 ) . The graph of the function is positive on ( − ∞ , − 7 )
Explanation
Understanding the Problem We are given a polynomial function with roots -7 (multiplicity 2), -1 (multiplicity 1), 2 (multiplicity 4), and 4 (multiplicity 1). The leading coefficient is positive, and the degree is even. We need to determine which statement about the graph is true.
Determining the Degree First, let's find the degree of the polynomial. The degree is the sum of the multiplicities of the roots: 2 + 1 + 4 + 1 = 8 . So, the polynomial is of degree 8, which is even.
General Form of the Polynomial The general form of the polynomial can be written as f ( x ) = a ( x + 7 ) 2 ( x + 1 ) ( x − 2 ) 4 ( x − 4 ) , where 0"> a > 0 is the leading coefficient.
Analyzing the Sign of f(x) Now, let's analyze the sign of f ( x ) on the given intervals:
On the interval ( 2 , 4 ) , we have:
0"> ( x + 7 ) 2 > 0
0"> ( x + 1 ) > 0
0"> ( x − 2 ) 4 > 0
( x − 4 ) < 0 Since 0"> a > 0 , the sign of f ( x ) is determined by the product of these factors. Thus, 0 \cdot > 0 \cdot > 0 \\\cdot < 0 = < 0"> f ( x ) > 0 ⋅ > 0 ⋅ > 0 ⋅ < 0 =< 0 . So, f ( x ) < 0 on ( 2 , 4 ) .
On the interval ( 4 , ∞ ) , we have:
0"> ( x + 7 ) 2 > 0
0"> ( x + 1 ) > 0
0"> ( x − 2 ) 4 > 0
0"> ( x − 4 ) > 0 Since 0"> a > 0 , the sign of f ( x ) is determined by the product of these factors. Thus, 0 \\cdot > 0 \\cdot > 0 \\cdot > 0 = > 0"> f ( x ) > 0 c d o t > 0 c d o t > 0 c d o t > 0 => 0 . So, 0"> f ( x ) > 0 on ( 4 , ∞ ) .
On the interval ( − ∞ , − 7 ) , we have:
0"> ( x + 7 ) 2 > 0
( x + 1 ) < 0
0"> ( x − 2 ) 4 > 0
( x − 4 ) < 0 Since 0"> a > 0 , the sign of f ( x ) is determined by the product of these factors. Thus, 0 \\cdot < 0 \\cdot > 0 \\cdot < 0 = > 0"> f ( x ) > 0 c d o t < 0 c d o t > 0 c d o t < 0 => 0 . So, 0"> f ( x ) > 0 on ( − ∞ , − 7 ) .
On the interval ( − 7 , − 1 ) , we have:
0"> ( x + 7 ) 2 > 0
0"> ( x + 1 ) > 0
0"> ( x − 2 ) 4 > 0
( x − 4 ) < 0 Since 0"> a > 0 , the sign of f ( x ) is determined by the product of these factors. Thus, 0 \\cdot > 0 \\cdot > 0 \\cdot < 0 = < 0"> f ( x ) > 0 c d o t > 0 c d o t > 0 c d o t < 0 =< 0 . So, f ( x ) < 0 on ( − 7 , − 1 ) .
Determining the True Statement Comparing our results with the given statements:
The graph of the function is positive on ( 2 , 4 ) . False
The graph of the function is negative on ( 4 , ∞ ) . False
The graph of the function is positive on ( − ∞ , − 7 ) . True
The graph of the function is negative on ( − 7 , − 1 ) . True
Since we need to choose only one true statement, and the third option is present, we choose that one.
Final Answer Therefore, the correct statement is: The graph of the function is positive on ( − ∞ , − 7 ) .
Examples
Understanding the behavior of polynomial functions is crucial in many real-world applications. For instance, engineers use polynomials to model the trajectory of a projectile, where the roots represent key points like launch and landing. Similarly, economists might use polynomials to model market trends, where understanding the intervals of positive and negative values helps predict periods of growth or decline. By analyzing the roots and their multiplicities, we can gain insights into the overall behavior of the system being modeled, allowing for better predictions and decision-making. For example, the polynomial f ( x ) = ( x + 2 ) 2 ( x − 1 ) has roots at x = − 2 (multiplicity 2) and x = 1 (multiplicity 1). This could represent a scenario where a company's profit is modeled by f ( x ) , with x being the investment. The root x = − 2 indicates a break-even point, and the multiplicity of 2 suggests a stable point. The root x = 1 indicates another break-even point, after which the profit becomes positive.
