The problem asks for the points at which the tangent line to the graph of the function is horizontal.

The function given is:

$y = \frac{1}{3}x^3 - 2x + 4$

To find the points where the tangent line is horizontal, we need to find where the derivative of the function equals zero. The derivative $y'$ represents the slope of the tangent line, and a horizontal line has a slope of 0. So, we proceed as follows:

1. Find the derivative of the function:

$y' = \frac{d}{dx} \left( \frac{1}{3}x^3 - 2x + 4 \right)$

$y' = x^2 - 2$

2. Set the derivative equal to zero to find critical points:

$x^2 - 2 = 0$

$x^2 = 2$

$x = \pm \sqrt{2}$

3. Find the corresponding y-values:

For $x = \sqrt{2}$:

$y = \frac{1}{3}(\sqrt{2})^3 - 2(\sqrt{2}) + 4 = \frac{1}{3}(2\sqrt{2}) - 2\sqrt{2} + 4 = \frac{2\sqrt{2}}{3} - 2\sqrt{2} + 4$

Simplifying:

$y = \frac{2\sqrt{2}}{3} - \frac{6\sqrt{2}}{3} + 4 = \frac{-4\sqrt{2}}{3} + 4$

For $x = -\sqrt{2}$:

$y = \frac{1}{3}(-\sqrt{2})^3 - 2(-\sqrt{2}) + 4 = \frac{1}{3}(-2\sqrt{2}) + 2\sqrt{2} + 4$

Simplifying:

$y = \frac{-2\sqrt{2}}{3} + \frac{6\sqrt{2}}{3} + 4 = \frac{4\sqrt{2}}{3} + 4$

Conclusion:

The points at which the tangent line is horizontal are:

1. $\left(\sqrt{2}, \frac{-4\sqrt{2}}{3} + 4\right)$
2. $\left(-\sqrt{2}, \frac{4\sqrt{2}}{3} + 4\right)$

Would you like more details or any clarifications on these calculations?

Here are five related questions you might find useful:

1. What is the significance of the second derivative in this context?
2. How can you use the first derivative test to classify these critical points?
3. What does the shape of the graph around these points look like?
4. Can horizontal tangents indicate maximum or minimum points on a graph?
5. How does this method change if the function had higher-order terms?

Tip: Horizontal tangents often occur at local maxima, minima, or points of inflection. Always analyze the context further with second derivatives or first derivative tests.