Given the quadratic function \( y = 4x - x^2 \), we can find the position of the maximum or minimum using calculus, following a similar procedure as before.

### Step 1: Express the quadratic function
The quadratic function is:

\[
y = -x^2 + 4x
\]

This is in the standard form \( y = ax^2 + bx + c \), where:
- \( a = -1 \)
- \( b = 4 \)
- \( c = 0 \)

### Step 2: Find the first derivative
To find the critical points, take the first derivative of \( y \) with respect to \( x \):

\[
\frac{dy}{dx} = \frac{d}{dx}(-x^2 + 4x)
\]

Applying the power rule:

\[
\frac{dy}{dx} = -2x + 4
\]

### Step 3: Set the first derivative to zero
Set the first derivative equal to zero and solve for \( x \):

\[
-2x + 4 = 0
\]

\[
2x = 4
\]

\[
x = 2
\]

### Step 4: Identify the nature of the critical point
To determine whether this critical point is a maximum or minimum, find the second derivative:

\[
\frac{d^2y}{dx^2} = \frac{d}{dx}(-2x + 4) = -2
\]

Since the second derivative is negative (\( -2 < 0 \)), the function is concave down, indicating that the critical point \( x = 2 \) is a **maximum**.

### Step 5: Find the corresponding \( y \)-coordinate
Substitute \( x = 2 \) back into the original function to find the maximum value:

\[
y = -x^2 + 4x = -(2)^2 + 4(2)
\]

\[
y = -4 + 8 = 4
\]

### Conclusion
The maximum value of the function \( y = 4x - x^2 \) occurs at \( x = 2 \), and the maximum value is \( y = 4 \).

---

Would you like further details or have any questions?

Here are some related questions you might find interesting:

1. How would you graph the function \( y = 4x - x^2 \)?
2. What is the significance of the vertex in the context of this problem?
3. How does the quadratic function relate to physical problems like projectile motion?
4. What are the intercepts of the function \( y = 4x - x^2 \)?
5. Can you express the function \( y = 4x - x^2 \) in vertex form?
6. What happens to the maximum value if the coefficient of \( x^2 \) changes?
7. How would the graph change if the function were \( y = x^2 - 4x \)?
8. Can the function \( y = 4x - x^2 \) have any other critical points?

**Tip:** When dealing with quadratic functions, the vertex is always a critical point, and it gives the maximum or minimum value depending on the sign of the leading coefficient.

Learn how to find the maximum value of the quadratic function y = 4x - x^2 using calculus. Discover step-by-step solutions, including finding critical points, applying the second derivative test, and determining concavity. Suitable for advanced high school students and calculus learners.

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