To analyze the function $y = 4x - x^2$, let's follow the steps indicated in your task:

Step 1: Find critical points

Critical points occur where the derivative $y'$ is zero or undefined.

1. Compute the derivative: $y' = \frac{d}{dx}(4x - x^2) = 4 - 2x$
2. Set the derivative equal to zero: $4 - 2x = 0 \implies x = 2$ The critical point is at $x = 2$.

Step 2: Determine maximum and minimum points

To classify the critical point $x = 2$, we analyze the second derivative:

1. Compute the second derivative: $y'' = \frac{d}{dx}(4 - 2x) = -2$
2. Since $y'' < 0$, the critical point $x = 2$ is a local maximum.

To find the corresponding $y$-value: $y(2) = 4(2) - (2)^2 = 8 - 4 = 4$ The local maximum is at $(2, 4)$.

Step 3: Find inflection points

Inflection points occur where $y'' = 0$ or changes sign. Here: $y'' = -2 \neq 0$ Since $y''$ is constant and negative, there are no inflection points.

Step 4: Find intervals of increase and decrease

1. Analyze $y' = 4 - 2x$:
   • For $y' > 0$: $4 - 2x > 0 \implies x < 2$ (increasing interval).
   • For $y' < 0$: $4 - 2x < 0 \implies x > 2$ (decreasing interval).

Thus:

• Increasing: $(-\infty, 2)$
• Decreasing: $(2, \infty)$

Step 5: Graph the function

1. Plot the vertex (local maximum) at $(2, 4)$.
2. Note that the function is a downward-opening parabola since the coefficient of $x^2$ is negative ($-1$).

Would you like me to provide a plot of the graph? Let me know if you have any specific questions!

Relative Questions:

1. How are critical points used in determining the behavior of a function?
2. What is the significance of the second derivative in classifying critical points?
3. How can you confirm whether a point is a local minimum, maximum, or neither?
4. How does the sign of the leading coefficient of $x^2$ affect the graph's shape?
5. Can there be more than one critical point for a quadratic function?

Tip:

Always check the concavity of the function using the second derivative to confirm the nature of a critical point.