Which of the following graphs could be the graph of the function $f(x)=0.03 x^2(x^2-25)$?

The function is a polynomial of degree 4 with a positive leading coefficient, so it opens upwards on both ends.
The function has roots at x = 0 (multiplicity 2) and x = ± 5 (multiplicity 1).
The function is even, symmetric with respect to the y-axis.
The function has local minima at x = ± 12.5 ​ with a value of approximately -4.6875, and a local maximum at x = 0 .
The graph that matches these characteristics is the one that touches the x-axis at 0 and crosses at ± 5 , with minima below the x-axis. Therefore, the answer is ​ (The graph that satisfies the above conditions).

Explanation

Analyze the function We are given the function f ( x ) = 0.03 x 2 ( x 2 − 25 ) and asked to identify its graph. Let's analyze the function to determine its key features.

Determine the end behavior First, we can rewrite the function as f ( x ) = 0.03 x 4 − 0.75 x 2 . This is a polynomial of degree 4. Since the leading coefficient (0.03) is positive, the end behavior of the function is that f ( x ) approaches + ∞ as x approaches both + ∞ and − ∞ .

Find the roots Next, we find the roots of the function by setting f ( x ) = 0 : 0.03 x 2 ( x 2 − 25 ) = 0 This gives us x 2 = 0 or x 2 − 25 = 0 . Thus, the roots are x = 0 (with multiplicity 2) and x = ± 5 (each with multiplicity 1).

Determine the symmetry Now, let's consider 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 and symmetric with respect to the y-axis.

Find the critical points To find the critical points, we take the derivative of f ( x ) : f ′ ( x ) = 0.12 x 3 − 1.5 x Setting f ′ ( x ) = 0 , we get: 0.12 x 3 − 1.5 x = 0 ⟹ x ( 0.12 x 2 − 1.5 ) = 0 So x = 0 or 0.12 x 2 = 1.5 ⟹ x 2 = 0.12 1.5 ​ = 12 150 ​ = 2 25 ​ = 12.5 . Thus, x = ± 12.5 ​ = ± 2 ​ 5 ​ = ± 2 5 2 ​ ​ ≈ ± 3.54 .

Evaluate the function at critical points Now we evaluate the function at the critical points to find the local extrema: f ( 0 ) = 0 , and f ( ± 12.5 ​ ) = 0.03 ( 12.5 ) ( 12.5 − 25 ) = 0.03 ( 12.5 ) ( − 12.5 ) = − 0.03 ( 12.5 ) 2 = − 0.03 ( 156.25 ) = − 4.6875 .

Identify the graph Based on the end behavior, roots, symmetry, and critical points, we can deduce the shape of the graph. The graph is symmetric with respect to the y-axis, has roots at x = − 5 , 0 , 5 , touches the x-axis at x = 0 , crosses the x-axis at x = − 5 and x = 5 , and has local minima at x = ± 12.5 ​ with f ( ± 12.5 ​ ) = − 4.6875 . Therefore, the graph should resemble a 'W' shape with a flattened bottom at the origin.


Examples
Understanding the behavior of polynomial functions like f ( x ) = 0.03 x 2 ( x 2 − 25 ) is crucial in various real-world applications. For instance, engineers use polynomial functions to model the stability of bridges or the trajectory of projectiles. By analyzing the roots and extrema of these functions, they can predict potential failure points or optimize performance. Similarly, economists use polynomial models to forecast market trends, where the roots might represent equilibrium points and the extrema indicate periods of maximum or minimum economic activity. This knowledge helps in making informed decisions and managing risks effectively.

Answered by GinnyAnswer | 2025-07-08

The graph of the function f ( x ) = 0.03 x 2 ( x 2 − 25 ) will open upwards, have roots at x = 0 (touching the axis) and x = \form − 5 and x = 5 (crossing the axis), and display local minima below the x-axis. This graph is symmetric with respect to the y-axis and resembles a 'W' shape. The selected graph option would be one that touches the x-axis at x = 0 and crosses at x = 5 and x = − 5 .
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Answered by Anonymous | 2025-07-13