To describe the behavior of the function $f(x) = -3x^4 + 5x^2 - 1$, let's analyze its key characteristics using proper mathematical notation:

1. End Behavior:

Since the leading term is $-3x^4$, a negative coefficient with an even degree indicates that both ends of the graph will point downward.

$\lim_{x \to \infty} f(x) = -\infty \quad \text{and} \quad \lim_{x \to -\infty} f(x) = -\infty$

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2. Intercepts:

• Y-intercept: $f(0) = -3(0)^4 + 5(0)^2 - 1 = -1$ The y-intercept is $(0, -1)$.

• X-intercepts: Solve: $-3x^4 + 5x^2 - 1 = 0$ Use substitution $u = x^2$, giving: $-3u^2 + 5u - 1 = 0$ Solving using the quadratic formula: $u = \frac{-5 \pm \sqrt{25 - 4(-3)(-1)}}{2(-3)} = \frac{-5 \pm \sqrt{13}}{-6}$ Thus: $x^2 = \frac{5 \pm \sqrt{13}}{6}$ The x-intercepts are: $x = \pm \sqrt{\frac{5 + \sqrt{13}}{6}} \quad \text{and} \quad x = \pm \sqrt{\frac{5 - \sqrt{13}}{6}}$

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3. Symmetry:

Since all exponents are even, the function is even, meaning: $f(-x) = f(x)$ The graph is symmetric about the y-axis.

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4. Critical Points:

Find the derivative: $f'(x) = -12x^3 + 10x$ Setting it to zero: $-12x^3 + 10x = 0$ Factor: $x(-12x^2 + 10) = 0$ Solving: $x = 0 \quad \text{or} \quad x = \pm \sqrt{\frac{5}{6}}$

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5. Concavity:

Find the second derivative: $f''(x) = -36x^2 + 10$ Solving $f''(x) = 0$: $-36x^2 + 10 = 0 \implies x^2 = \frac{5}{18}$ Points of inflection occur at: $x = \pm \sqrt{\frac{5}{18}}$

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Would you like further explanations on any of these points?

5 Follow-Up Questions:

1. What is the significance of even and odd functions in graphing?
2. How do you determine points of inflection using the second derivative?
3. Why do even-degree polynomials have similar end behaviors on both sides?
4. How can you solve higher-degree polynomials using substitution?
5. What role do critical points play in determining function behavior?

Tip:
When analyzing a polynomial function, always start by examining the leading term to predict end behavior.