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Chapter 3: Problem 2

These exercises review topics covered in earlier sections. The concepts are used in solving the applications that follow in this exercise set. Do not use a calculator. Find the maximum y-value on the graph of \(y=-2 x^{2}+8 x-5\).

Short Answer

Expert verified
The maximum y-value is 3.

Step by step solution

01

Identify the Type of Function

The given function is a quadratic function of the form \(y = ax^2 + bx + c\). Here, \(a = -2\), \(b = 8\), and \(c = -5\). It is a parabola, and since the coefficient of \(x^2\) is negative, it opens downwards.
02

Use the Formula for Vertex

The vertex of a quadratic function \(y = ax^2 + bx + c\) is given by the formula \(x = -\frac{b}{2a}\). This gives us the x-coordinate of the vertex, which, for a downward-opening parabola, yields the maximum y-value.
03

Calculate the x-coordinate of the Vertex

Substitute the values of \(a\) and \(b\) into the vertex formula: \(-\frac{8}{2(-2)} = -\frac{8}{-4} = 2\). The x-coordinate of the vertex is \(x = 2\).
04

Calculate the Maximum y-value

Substitute \(x = 2\) back into the original equation to find the maximum y-value: \(y = -2(2)^2 + 8(2) - 5\). This simplifies to \(y = -2(4) + 16 - 5 = -8 + 16 - 5 = 3\).
05

Confirm the Result

Since the parabola opens downwards and \(x=2\) is the vertex, \(y = 3\) is indeed the maximum y-value of the function. This result is confirmed by ensuring all calculations follow the method for finding maximum values of quadratic functions.

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Key Concepts

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

Parabola
A quadratic function is represented as a parabola on a graph. It is a smooth, symmetrical curve that can either open upwards or downwards.
  • If the coefficient of the quadratic term (\(x^2\)) is positive, the parabola opens upwards.
  • If this coefficient is negative, like in the function \(y = -2x^2 + 8x - 5\), the parabola opens downwards.
The direction in which a parabola opens affects the nature of its extremum:
  • Upward-opening parabolas have a minimum value.
  • Downward-opening parabolas, like our example, have a maximum value.
Understanding these characteristics is crucial when determining the maximum or minimum values of a quadratic function without using graphical tools.
Vertex Formula
The vertex of a parabola is a significant point that represents either its maximum or minimum value. For parabola given by the quadratic function \(y = ax^2 + bx + c\), the vertex's x-coordinate can be found using the formula:\[x = -\frac{b}{2a}\]This formula is derived from completing the square or using calculus principles. In our function \(y = -2x^2 + 8x - 5\), plugging in \(a = -2\) and \(b = 8\) into the formula gives us:\[x = -\frac{8}{2(-2)} = 2\]The x-coordinate of the vertex thus simplifies our task of finding the y-value at this point, leading us directly to the vertex, which is key for calculating maximum or minimum values.
Maximum Value
Once the x-coordinate of the vertex is determined, finding the corresponding y-coordinate gives us the maximum value for parabolas that open downward.
By substituting the x-value back into the function \(y = -2x^2 + 8x - 5\):
  • At \(x = 2\)
  • Substitute: \(y = -2(2)^2 + 8 \times 2 - 5\)
  • Calculate: \(y = -8 + 16 - 5\)
  • Result: \(y = 3\)
This calculation shows that \(y = 3\) is the maximum value of the function. For downward-opening parabolas, this y-value is the highest point on the graph, highlighting the parabola's highest elevation point, providing a practical approach to identify maxima without plotting the function.

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