Let's work through the given function $f(x) = x^4 - 16x^2$ and address the steps requested:

a) Factoring to find the x-intercepts:

The function can be factored as: $f(x) = x^2(x^2 - 16) = x^2(x - 4)(x + 4)$ Setting $f(x) = 0$ to find the x-intercepts: $x^2(x - 4)(x + 4) = 0$ This gives the solutions: $x = 0, \, x = 4, \, x = -4$ These are the x-intercepts.

b) Behavior of the graph:

• The graph crosses the x-axis at $x = 4$ and $x = -4$. This happens because at these points the factors $(x - 4)$ and $(x + 4)$ have a power of 1 (odd power), meaning the sign of the function changes.
• The graph touches the x-axis and turns around at $x = 0$. This occurs because the factor $x^2$ has an even power, so the function does not change sign here, resulting in a touch and turn around at this point.

c) Finding the y-intercept:

The y-intercept is found by calculating $f(0)$: $f(0) = 0^4 - 16(0^2) = 0$ Thus, the y-intercept is $f(0) = 0$.

Would you like more details on any of these points?

Here are 5 related questions for further exploration:

1. How do we classify the behavior of the graph at points where the function crosses versus touches the x-axis?
2. What is the significance of the multiplicity of roots in polynomial functions?
3. Can you explain how to determine the end behavior of $f(x) = x^4 - 16x^2$?
4. How would the graph change if we modified the function to $f(x) = x^4 - 16x^2 + 1$?
5. What are some other methods for solving polynomial equations beyond factoring?

Tip: Always look at the degree and leading coefficient of a polynomial to predict its end behavior.