Which of the following graphs could be the graph of the function [tex]$f(x)=0.03 x^2(x^2-25)$[/tex]?
Answer (2)
The function f ( x ) = 0.03 x 2 ( x 2 − 25 ) has roots at x = − 5 , 0 , 5 , opens upwards, and is symmetric about the y-axis. The graph touches the x-axis at x = 0 and crosses at x = − 5 and x = 5 . Look for a graph that matches these characteristics.
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Find the roots of the function: x = 0 (multiplicity 2), x = ± 5 .
Determine the end behavior: Opens upwards on both ends.
Check the symmetry: The function is even, symmetric with respect to the y-axis.
Analyze behavior around roots: Touches x-axis at x = 0 , crosses at x = ± 5 . The graph has roots at x = − 5 , 0 , 5 , touches the x-axis at x = 0 , crosses the x-axis at x = ± 5 , opens upwards on both ends, and is symmetric with respect to the y-axis.
The function values at x = 1 and x = − 1 are f ( 1 ) = f ( − 1 ) = − 0.72 .
The graph of the function is f ( x ) = 0.03 x 2 ( x 2 − 25 ) .
Explanation
Analyzing the Function We are given the function f ( x ) = 0.03 x 2 ( x 2 − 25 ) and we need to identify its graph. Let's analyze the function to determine its key features.
Finding the Roots First, let's find the roots of the function. The roots are the values of x for which f ( x ) = 0 . So, we have 0.03 x 2 ( x 2 − 25 ) = 0 This gives us x 2 = 0 or x 2 − 25 = 0 . Thus, x = 0 (with multiplicity 2) and x = ± 5 .
Determining End Behavior Next, let's determine the end behavior of the function. Since the function is a polynomial of degree 4 with a positive leading coefficient (0.03), as x approaches ± ∞ , f ( x ) approaches + ∞ . This means the graph opens upwards on both ends.
Checking Symmetry Now, let's check the symmetry of the function. We have f ( − x ) = 0.03 ( − x ) 2 (( − x ) 2 − 25 ) = 0.03 x 2 ( x 2 − 25 ) = f ( x ) Since f ( − x ) = f ( x ) , the function is even, which means the graph is symmetric with respect to the y-axis.
Finding the y-intercept The y-intercept of the function is the value of f ( x ) when x = 0 . In this case, f ( 0 ) = 0.03 ( 0 ) 2 ( 0 2 − 25 ) = 0 . So, the y-intercept is 0.
Analyzing Behavior Around Roots Now, let's analyze the behavior of the function around the roots. At x = 0 , the function has a root with multiplicity 2, which means the graph touches the x-axis at x = 0 but does not cross it. At x = ± 5 , the function has simple roots, which means the graph crosses the x-axis at these points.
Sketching the Graph Based on the above information, we can sketch the graph. The graph should:
Have roots at x = − 5 , 0 , 5 .
Touch the x-axis at x = 0 and cross it at x = ± 5 .
Open upwards on both ends.
Be symmetric with respect to the y-axis.
The function values at x = 1 and x = − 1 are f ( 1 ) = f ( − 1 ) = 0.03 ( 1 ) ( 1 − 25 ) = 0.03 ( − 24 ) = − 0.72 .
Conclusion The graph of the function f ( x ) = 0.03 x 2 ( x 2 − 25 ) has roots at x = − 5 , 0 , 5 , touches the x-axis at x = 0 , crosses the x-axis at x = ± 5 , opens upwards on both ends, and is symmetric with respect to the y-axis.
Examples
Understanding the behavior of polynomial functions like f ( x ) = 0.03 x 2 ( x 2 − 25 ) is crucial in various fields such as physics and engineering. For instance, when designing a bridge, engineers need to analyze the load distribution, which can often be modeled by polynomial functions. The roots of the function represent critical points where the load is zero, and the shape of the graph helps visualize how the load is distributed across the structure. By understanding the end behavior and symmetry, engineers can ensure the bridge's stability and safety under different loading conditions. Similarly, in physics, analyzing the trajectory of a projectile involves understanding polynomial functions, where the roots represent the points where the projectile hits the ground, and the shape of the graph describes its path.
