To trace the curve given by \( y = x^4 - 8x^2 + 16 \), let's break the problem down step by step, examining its key features such as symmetry, critical points, and behavior at the limits.

### 1. **Symmetry**:
The function \( y = x^4 - 8x^2 + 16 \) is even because all powers of \( x \) are even. Therefore, the curve is symmetric with respect to the y-axis. This means that the curve on the right side of the y-axis (for \( x \geq 0 \)) is a mirror image of the left side (for \( x \leq 0 \)).

### 2. **Critical Points**:
To find the critical points, we calculate the first derivative of \( y \) with respect to \( x \):

\[
\frac{dy}{dx} = 4x^3 - 16x
\]

To find the critical points, set \( \frac{dy}{dx} = 0 \):

\[
4x^3 - 16x = 0
\]

Factor the equation:

\[
4x(x^2 - 4) = 0
\]

\[
4x(x - 2)(x + 2) = 0
\]

Thus, the critical points are \( x = 0 \), \( x = 2 \), and \( x = -2 \).

### 3. **Second Derivative Test**:
To determine the nature of these critical points (whether they are maxima, minima, or points of inflection), we compute the second derivative:

\[
\frac{d^2y}{dx^2} = 12x^2 - 16
\]

- At \( x = 0 \):
  \[
  \frac{d^2y}{dx^2} = 12(0)^2 - 16 = -16
  \]
  Since \( \frac{d^2y}{dx^2} < 0 \), \( x = 0 \) is a **local maximum**.

- At \( x = 2 \) and \( x = -2 \):
  \[
  \frac{d^2y}{dx^2} = 12(2)^2 - 16 = 32 - 16 = 16
  \]
  Since \( \frac{d^2y}{dx^2} > 0 \), both \( x = 2 \) and \( x = -2 \) are **local minima**.

### 4. **Behavior at Infinity**:
To understand the behavior of the curve as \( x \to \infty \) and \( x \to -\infty \), look at the leading term \( x^4 \) in the function. As \( x \to \infty \) or \( x \to -\infty \), the term \( x^4 \) dominates, and \( y \to \infty \). Therefore, the curve rises to infinity on both ends.

### 5. **Y-Intercept**:
The y-intercept occurs when \( x = 0 \):

\[
y(0) = (0)^4 - 8(0)^2 + 16 = 16
\]

Thus, the curve crosses the y-axis at the point \( (0, 16) \).

### 6. **X-Intercepts**:
To find the x-intercepts, we set \( y = 0 \) and solve for \( x \):

\[
x^4 - 8x^2 + 16 = 0
\]

This is a quadratic in \( x^2 \). Let \( z = x^2 \), so the equation becomes:

\[
z^2 - 8z + 16 = 0
\]

Solve using the quadratic formula:

\[
z = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(16)}}{2(1)} = \frac{8 \pm \sqrt{64 - 64}}{2} = \frac{8 \pm 0}{2} = 4
\]

So, \( z = 4 \), which means \( x^2 = 4 \), or \( x = \pm 2 \).

Thus, the curve crosses the x-axis at the points \( (-2, 0) \) and \( (2, 0) \).

### 7. **Summary of the Curve**:
- **Symmetry**: Even function, symmetric about the y-axis.
- **Critical Points**: \( x = 0 \) is a local maximum, \( x = \pm 2 \) are local minima.
- **Intercepts**: 
  - Y-intercept: \( (0, 16) \)
  - X-intercepts: \( (-2, 0) \), \( (2, 0) \)
- **Behavior at infinity**: \( y \to \infty \) as \( x \to \infty \) or \( x \to -\infty \).
  
This provides a clear sketch of the curve's general shape.

---

Would you like further details or have any questions about this analysis? 

Here are 5 follow-up questions for you:
1. How do you apply the second derivative test to identify inflection points?
2. Can you describe how the curve behaves near \( x = 0 \)?
3. What role does symmetry play in simplifying curve tracing?
4. How does factoring help in finding the critical points of a function?
5. What does the leading term tell you about the behavior of a polynomial at large values of \( x \)?

**Tip**: When solving higher-degree polynomials, factoring or substitution (like \( z = x^2 \)) often simplifies the work significantly.

Learn how to trace the curve of the polynomial function y = x^4 - 8x^2 + 16. This solution covers key calculus concepts such as finding critical points using derivatives, applying the second derivative test, identifying symmetry, and determining x- and y-intercepts. Suitable for Grades 11-12.

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