Question

Graph the function y = 4x4 – 8x2 + 4. Which lists all of the turning points of the graph?

(0, 4)
(–1, 0) and (1, 0)
(–1, 0), (0, 4), and (1, 0)
(–4, 0), (–1, 0), (0, 4), and (1, 0)

Answer 1

To find the turning points of the function \( y = 4x^4 - 8x^2 + 4 \), we need to calculate its derivative and set it equal to zero to find the critical points.

1. **Find the derivative**:
\[
y' = \frac{d}{dx}(4x^4 - 8x^2 + 4) = 16x^3 - 16x
\]

2. **Set the derivative equal to zero**:
\[
16x^3 - 16x = 0
\]
Factor out the common term:
\[
16x(x^2 - 1) = 0
\]
This gives us:
\[
16x(x-1)(x+1) = 0
\]

3. **Solve for \( x \)**:
\[
x = 0, \quad x = 1, \quad x = -1
\]

4. **Find the corresponding \( y \)-values at these \( x \)-values**:
- For \( x = 0 \):
\[
y(0) = 4(0)^4 - 8(0)^2 + 4 = 4
\]
- For \( x = 1 \):
\[
y(1) = 4(1)^4 - 8(1)^2 + 4 = 4 - 8 + 4 = 0
\]
- For \( x = -1 \):
\[
y(-1) = 4(-1)^4 - 8(-1)^2 + 4 = 4 - 8 + 4 = 0
\]

5. **Summarize the turning points**:
From the calculations above, the turning points are:
- \( (0, 4) \)
- \( (1, 0) \)
- \( (-1, 0) \)

Thus, the turning points of the graph are:

**(–1, 0), (0, 4), and (1, 0)**

The correct option is:
**(–1, 0), (0, 4), and (1, 0)**.