Problem 41 - Algebra and Trigonometry

Question

Find the maximum or minimum value of the function. $$ f(x)=3-x-\frac{1}{2} x^{2} $$

Step-by-Step Solution

Answer

The maximum value of the function is -1.

Step 1: Determine the Type of Function

The given function is a quadratic function of the form \[ f(x) = ax^2 + bx + c \]where in this case, \( a = -\frac{1}{2} \), \( b = -1 \), and \( c = 3 \). Because \( a < 0 \), the parabola opens downwards, indicating the function has a maximum value at its vertex.

Step 2: Find the Vertex Using the Vertex Formula

The vertex of a quadratic function \( ax^2 + bx + c \) is at \( x = -\frac{b}{2a} \). Substituting \( a = -\frac{1}{2} \) and \( b = -1 \) into the formula gives:\[ x = -\frac{-1}{2(-\frac{1}{2})} = -\frac{1}{-1} = 2 \] Thus, the x-coordinate of the vertex is \( x = 2 \).

Step 3: Calculate the Value of the Function at the Vertex

Substitute \( x = 2 \) into the function to find the maximum value:\[ f(2) = 3 - 2 - \frac{1}{2}(2)^2 \]\[ = 3 - 2 - \frac{1}{2}(4) \]\[ = 3 - 2 - 2 \]\[ = -1 \]Thus, the maximum value of the function is -1.

Key Concepts

Maximum Value

In the realm of quadratic functions, understanding the concept of maximum and minimum values is pivotal. These values provide vital information about the behavior of a function. When we talk about the maximum value of a quadratic function, we refer to the highest point on its graph. This is crucial, especially in cases related to scenarios like finding the peak of a projectile's trajectory.Quadratic functions can be expressed generically in the form:\[ f(x) = ax^2 + bx + c \]where:

• \( a \), \( b \), \( c \) are constants,
• \( a eq 0 \) since a quadratic is characterized by the term \( ax^2 \).

If the coefficient \( a \) is negative, the parabola formed by the function opens downwards. This implies that the function will have a maximum value at its vertex, which is the peak point. In such cases, determining the vertex is key to finding this maximum value.

Vertex of a Parabola

The vertex of a parabola formed by a quadratic function is a point of special interest. It represents the turning point, which is either the maximum or minimum value of the function, depending on the direction the parabola opens.To find the vertex, we use the vertex formula for a quadratic equation given by \( ax^2 + bx + c \):\[ x = -\frac{b}{2a} \]Substituting the values of \( a \) and \( b \) from the given function \( f(x) = 3 - x - \frac{1}{2}x^2 \) leads us to the x-coordinate of the vertex:

• \( a = -\frac{1}{2} \)
• \( b = -1 \)
• Thus, \( x = -\frac{-1}{2(-\frac{1}{2})} = 2 \)

At \( x = 2 \), we find the function's vertex. This x-coordinate can be plugged back into the function to find the corresponding y-value, thus defining the vertex completely as the point (2, \( f(2) \)).

Quadratic Formula

The quadratic formula is often used to find the roots (or zeros) of a quadratic function, but it's also indirectly related to finding maximum or minimum values through understanding the vertex.The formula itself is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]While the primary use is to solve for \( x \) when \( f(x) = 0 \), being familiar with this formula is integral to comprehending the complete analysis of quadratic functions. It provides insights into the points where the quadratic graph intersects the x-axis. However, for tackling problems centered on maximum value as in our given exercise, the vertex formula is more directly applied.This broader understanding of quadratic functions through formulas and concepts helps demystify how these functions behave, whether finding where they meet axes or determining highest and lowest points on their curves.