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Chapter 8: Problem 115

Determine whether \(f(x)=-3 x^{2}+120 x+50\) has a maximum or a minimum value, and then find the value.

Short Answer

Expert verified
The function has a maximum value of 1250.

Step by step solution

01

Identify the Type of Parabola

Given the quadratic function is of the form \[ f(x) = ax^2 + bx + c \]. Identify the coefficient \( a = -3 \), which is negative. When \( a \) is negative, the parabola opens downwards, indicating a maximum value.
02

Find the Vertex

For a quadratic function \( f(x) = ax^2 + bx + c \), the vertex \( (h, k) \) can be found using \( h = -\frac{b}{2a} \). Substitute \( a = -3 \) and \( b = 120 \) into the formula: \[ h = -\frac{120}{2(-3)} = 20 \].
03

Calculate the Maximum Value

Substitute \( x = 20 \) back into the function \( f(x) \) to find the maximum value: \[ f(20) = -3(20)^2 + 120(20) + 50 \]. Simplify: \[ f(20) = -3(400) + 2400 + 50 = -1200 + 2400 + 50 = 1250 \]. So, the maximum value is 1250.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

maximum value of quadratic function
Quadratic functions can either have a maximum or a minimum value based on the direction in which the parabola opens. This is determined by the coefficient 'a' in the function of the form \(f(x) = ax^2 + bx + c\).
  • If \(a > 0\), the parabola opens upwards, resulting in a minimum value at the vertex.
  • If \(a < 0\), the parabola opens downwards, resulting in a maximum value at the vertex.


In the given problem, \(f(x) = -3x^2 + 120x + 50\), we have \(a = -3\).
Since \(-3 < 0\), the parabola opens downwards, and thus the function has a maximum value.
vertex of a parabola
The vertex of a parabola is the highest or lowest point of the curve, depending on the direction it opens. It is a crucial point where the maximum or minimum value of the function is achieved.
To find the vertex \((h, k)\) of a quadratic function \(f(x) = ax^2 + bx + c\):
  • Calculate the x-coordinate, \(h\), using the formula \(h = -\frac{b}{2a}\).
  • Substitute \(\text{a}\) and \(\text{b}\) from the quadratic function into this formula.

For the given function, \(a = -3\) and \(b = 120\):
\(h = -\frac{120}{2(-3)} = 20\).
The x-coordinate of the vertex is 20.
To find the y-coordinate \((k)\), substitute \(h \) back into the function:
\f(h) = -3(20)^2 + 120(20) + 50\>
Simplifying this,
\f(20) = -3(400) + 2400 + 50 = 1250\.
Therefore, the vertex is \( (20, 1250)\).
direction of a parabola
The direction of a parabola describes whether it opens upwards or downwards. This is determined by the leading coefficient \(a\) in the quadratic expression \(f(x) = ax^2 + bx + c\).
  • When \((a > 0)\), the parabola opens upwards, like a 'U' shape.
  • When \(a < 0\), the parabola opens downwards, like an upside-down 'U'.

Identifying the direction helps in predicting the nature of the function's vertex:
  • For parabolas that open upwards, the vertex is the minimum point.
  • For parabolas that open downwards, the vertex is the maximum point.

Given the function \(f(x) = -3x^2 + 120x + 50\), with \(a = -3\), we can see that the parabola opens downwards, allowing us to determine the vertex as the point of the maximum value.

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