Math Problem Statement

Match the polynomial function with its graph. f(x) = − 1/4 x exponent 4 + 3x2 sqare

Solution

To match the polynomial function f(x)=14x4+3x2f(x) = -\frac{1}{4}x^4 + 3x^2 with its graph, let's analyze the function step by step:

1. Degree and End Behavior:

  • The degree of the polynomial is 4 (the highest exponent is 4), which means it is a quartic function.
  • Since the leading term has a negative coefficient 14x4-\frac{1}{4}x^4, the ends of the graph will point downward. As xx \to \infty or xx \to -\infty, f(x)f(x) \to -\infty. This gives the graph a "W" shape, but inverted because of the negative leading term.

2. Turning Points:

  • A quartic polynomial can have up to 3 turning points (local maxima and minima). The number of turning points helps determine the general shape of the graph.

3. Symmetry:

  • The function is an even function because only even powers of xx are present (no odd powers of xx). Thus, the graph is symmetric with respect to the y-axis.

4. Intercepts:

  • y-intercept: Plugging x=0x = 0 into the function, we get: f(0)=14(0)4+3(0)2=0f(0) = -\frac{1}{4}(0)^4 + 3(0)^2 = 0 So the graph passes through the origin (0,0)(0, 0).
  • x-intercepts: To find the x-intercepts, solve f(x)=0f(x) = 0: 14x4+3x2=0-\frac{1}{4}x^4 + 3x^2 = 0 Factor the equation: x2(14x2+3)=0x^2\left(-\frac{1}{4}x^2 + 3\right) = 0 The solutions are: x2=0or14x2+3=0x^2 = 0 \quad \text{or} \quad -\frac{1}{4}x^2 + 3 = 0 From x2=0x^2 = 0, we get x=0x = 0. Solving 14x2+3=0-\frac{1}{4}x^2 + 3 = 0, we get: 14x2=3x2=12x=±23-\frac{1}{4}x^2 = -3 \quad \Rightarrow \quad x^2 = 12 \quad \Rightarrow \quad x = \pm 2\sqrt{3} So the x-intercepts are x=0x = 0 and x=±23x = \pm 2\sqrt{3}.

5. Behavior near the origin:

  • Near the origin, the graph behaves like a parabola because the 3x23x^2 term dominates. The 14x4-\frac{1}{4}x^4 term will flatten the graph away from the origin as x|x| increases.

Conclusion:

  • The graph should be symmetric about the y-axis, pass through the origin, have three x-intercepts (including ±23\pm 2\sqrt{3}), and exhibit "W"-like behavior (inverted) because of the negative quartic term.

Would you like me to generate a graph for this function or clarify any specific part?

Related Questions:

  1. What are the critical points of f(x)f(x) and how do we find them?
  2. How does the graph change if we alter the coefficient of the quartic term?
  3. Can you explain the end behavior of a general quartic function?
  4. How do we calculate turning points using derivatives for f(x)f(x)?
  5. What are the possible shapes of graphs for even-degree polynomials?

Tip:

The symmetry of a polynomial function can often be identified by checking whether all powers of xx are even (even function) or odd (odd function).

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Math Problem Analysis

Mathematical Concepts

Polynomial Functions
End Behavior
Symmetry
Intercepts
Turning Points

Formulas

f(x) = -\frac{1}{4}x^4 + 3x^2
End Behavior for quartic functions with a negative leading coefficient
x-intercepts using factorization

Theorems

Even function symmetry (functions with only even powers are symmetric about the y-axis)
Fundamental theorem of algebra (maximum number of real roots)
Turning points of polynomial functions (based on degree)

Suitable Grade Level

Grades 11-12 or early college level