Understanding the Maximum of the Quadratic Function $ P(x) = -x^2 + 4x + m $

When analyzing quadratic functions in the form $ P(x) = ax^2 + bx + c $, one of the most important concepts is identifying where the function reaches its maximum or minimum. In this case, we examine the downward-opening parabola defined by:

$$
P(x) = -x^2 + 4x + m
$$

Understanding the Context

Here, the coefficient $ a = -1 $, $ b = 4 $, and $ c = m $. Since $ a < 0 $, the parabola opens downward, meaning it has a maximum value at its vertex.

Finding the x-Coordinate of the Vertex

The $ x $-coordinate of the vertex of any quadratic function is given by the formula:

$$
x = rac{-b}{2a}
$$

Solution: The maximum of $ P(x) = -x^2 + 4x + m $ occurs at the vertex. The $ x $-coordinate of the vertex is $ x = \frac{-b}{2a} = \frac{-4}{-2} = 2 $. Substitute $ x = 2 $ into $ P(x) $:

Key Insights

Substituting $ a = -1 $ and $ b = 4 $:

$$
x = rac{-4}{2(-1)} = rac{-4}{-2} = 2
$$

So, the vertex occurs at $ x = 2 $, which is the point where the function $ P(x) $ reaches its maximum value.

Evaluating the Maximum Value by Substituting $ x = 2 $

To find the actual maximum value of $ P(x) $, substitute $ x = 2 $ into the expression:

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Final Thoughts

$$
P(2) = -(2)^2 + 4(2) + m = -4 + 8 + m = 4 + m
$$

Thus, the maximum value of $ P(x) $ is $ 4 + m $, occurring at $ x = 2 $.

Key Takeaways

  • The vertex of $ P(x) = -x^2 + 4x + m $ is at $ x = 2 $, the x-coordinate where the maximum occurs.
  • Evaluating the function at $ x = 2 $ yields the peak value: $ P(2) = 4 + m $.
  • Understanding the vertex form helps students and learners determine key features like maximums, minima, and symmetry in quadratic functions.

This insight is crucial not only for solving optimization problems but also for graphing and interpreting real-world scenarios modeled by quadratic functions.

By recognizing that the maximum of $ P(x) $ occurs at $ x = 2 $, and computing $ P(2) = 4 + m $, you gain a powerful tool for analyzing and visualizing quadratic behavior.