Chapter 4: Problem 29
Find the maximum or minimum value for each function (whichever is appropriate). State whether the value is a maximum or minimum. $$y=-2 x^{2}-3 x+2$$
Short Answer
Expert verified
The maximum value is \(-\frac{11}{8}\), at \(x = \frac{3}{4}\).
Step by step solution
01
Identify the Type of Quadratic Function
The function given is a quadratic function in the form of \(y = ax^2 + bx + c\). Here, \(a = -2\), \(b = -3\), and \(c = 2\). Since \(a < 0\), the parabola opens downwards, and the vertex represents a maximum point.
02
Find the Vertex Formula
The vertex of a parabola given by \(y = ax^2 + bx + c\) is at \(x = -\frac{b}{2a}\). Substituting \(a = -2\) and \(b = -3\) into the formula gives \(x = -\frac{-3}{2(-2)} = \frac{3}{4}\).
03
Calculate the Maximum Function Value at the Vertex
Substitute \(x = \frac{3}{4}\) back into the function to find \(y\): \[ y = -2\left( \frac{3}{4} \right)^2 - 3\left( \frac{3}{4} \right) + 2 \]\[ y = -2\left( \frac{9}{16} \right) - \frac{9}{4} + 2 \]\[ y = -\frac{9}{8} - \frac{18}{8} + \frac{16}{8} \]\[ y = -\frac{11}{8} \]Thus, the maximum value of the function is \(-\frac{11}{8}\).
04
Determine the Maximum Value
Since the function opens downwards and we have found the vertex, the value \(y = -\frac{11}{8}\) is indeed the maximum value of the function.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Maximum and Minimum Value
In quadratic functions, the maximum or minimum value is a crucial point, typically located at the vertex of the parabola. These values hint at the highest or lowest points that a function can achieve. Quadratic functions have standard forms of \( y = ax^2 + bx + c \). When nestled in the graph, these parabolas can either open upwards or downwards, dictated by the coefficient \( a \):
- If \( a > 0 \), the parabola opens upwards, and the vertex represents the minimum value of the function, because all values of \( y \) must be greater than this point.
- If \( a < 0 \), the parabola opens downwards, and the vertex represents the maximum value because all values of \( y \) must be less than this point.
Parabola Vertex
The vertex of a parabola is a fundamental point that serves as a peak or a trough of the function. It gives us both the x- and y-coordinates. The x-coordinate of a vertex in a quadratic function \( y = ax^2 + bx + c \) is found with the formula:\[x = -\frac{b}{2a}\]Substituting \( a = -2 \) and \( b = -3 \) into this formula yields:\[x = -\frac{-3}{2(-2)} = \frac{3}{4}\]This computation shows that the vertex's x-coordinate is \( \frac{3}{4} \). To find the corresponding y-value, substitute this x back into the original function:\[y = -2\left( \frac{3}{4} \right)^2 - 3\left( \frac{3}{4} \right) + 2\]Solving this gives us \( y = -\frac{11}{8} \). Thus, the vertex is at \( \left( \frac{3}{4}, -\frac{11}{8} \right) \), combining both coordinates into one pivotal point on the graph.
Downward-Opening Parabola
A downward-opening parabola is noted by its concave shape, which means it curves down like a hill. This shape contrasts with the upward-opening parabolas that resemble a valley. In quadratic terms, a downward-opening parabola arises when the coefficient \( a \) in \( y = ax^2 + bx + c \) is less than zero (\( a < 0 \)). This characteristic ensures that the parabola will never rise above the vertex, making the vertex strategically important as the maximum point of the function.Key features of a downward-opening parabola include:
- The extremum, or peak, is the highest point on the graph—this is your maximum for the function.
- The sides of the parabola will continue indefinitely towards negative infinity, never reaching or surpassing the y-coordinate of the vertex.
- Understanding the vertex's role and the direction of opening helps predict and analyze the behavior of such functions efficiently.