Solution: The quadratic $ p(t) = -t^2 + 14t + 30 $ opens downward. The vertex occurs at $ t = -\fracb2a = -\frac142(-1) = 7 $. Substituting $ t = 7 $, $ p(7) = -(7)^2 + 14(7) + 30 = -49 + 98 + 30 = 79 $. - Get link 4share
The Full Solution to Finding the Maximum Value of the Quadratic Function $ p(t) = -t^2 + 14t + 30 $
The Full Solution to Finding the Maximum Value of the Quadratic Function $ p(t) = -t^2 + 14t + 30 $
Understanding key features of quadratic functions is essential in algebra and real-world applications such as optimization and curve modeling. A particularly common yet insightful example involves analyzing the quadratic function $ p(t) = -t^2 + 14t + 30 $, which opens downward due to its negative leading coefficient. This analysis reveals both the vertex — the point of maximum value — and the function’s peak output.
Identifying the Direction and Vertex of the Quadratic
Understanding the Context
The given quadratic $ p(t) = -t^2 + 14t + 30 $ is in standard form $ ax^2 + bx + c $, where $ a = -1 $, $ b = 14 $, and $ c = 30 $. Because $ a < 0 $, the parabola opens downward, meaning it has a single maximum point — the vertex.
The $ t $-coordinate of the vertex is found using the formula $ t = -rac{b}{2a} $. Substituting the values:
$$
t = -rac{14}{2(-1)} = -rac{14}{-2} = 7
$$
This value, $ t = 7 $, represents the hour or moment when the quantity modeled by $ p(t) $ reaches its maximum.
Key Insights
Calculating the Maximum Value
To find the actual maximum value of $ p(t) $, substitute $ t = 7 $ back into the original equation:
$$
p(7) = -(7)^2 + 14(7) + 30 = -49 + 98 + 30 = 79
$$
Thus, the maximum value of the function is $ 79 $ at $ t = 7 $. This tells us that when $ t = 7 $, the system achieves its peak performance — whether modeling height, revenue, distance, or any real-world behavior described by this quadratic.
Summary
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- Function: $ p(t) = -t^2 + 14t + 30 $ opens downward ($ a < 0 $)
- Vertex occurs at $ t = -rac{b}{2a} = 7 $
- Maximum value is $ p(7) = 79 $
Knowing how to locate and compute the vertex is vital for solving optimization problems efficiently. Whether in physics, economics, or engineering, identifying such key points allows for precise modeling and informed decision-making — making the vertex a cornerstone of quadratic analysis.