Problem 41 - Algebra and Trigonometry

Question

An equation of a quadratic function is given. a. Determine, without graphing, whether the function has a minimum value or a maximum value. b. Find the minimum or maximum value and determine where it occurs. c. Identify the function's domain and its range. $$ f(x)=-4 x^{2}+8 x-3 $$

Step-by-Step Solution

Answer

The given function has a maximum value. The maximum value is 1, and it occurs at \(x = 1\). The domain of the function is all real numbers, \(x \in (-\infty, \infty)\), and the range is \(f(x) \in (-\infty, 1]\).

Step 1: Identify if the function has a minimum or maximum value

The coefficient of the \(x^{2}\) term in the given function is -4, which is less than 0. Therefore, the function opens downward and has a maximum value.

Step 2: Find the maximum value and where it occurs

The maximum value of the function occurs at the vertex of the parabola. The x-coordinate of the vertex can be found using the formula \(x = -\frac{b}{2a}\). Substituting \(a = -4\) and \(b = 8\), we get \(x = -\frac{8}{-8} = 1\). To find the maximum value, substitute \(x = 1\) into the function: \(f(1) = -4(1)^{2} + 8*1 - 3 = 1\). So, the maximum value is 1 and it occurs at \(x = 1\).

Step 3: Identify the function's domain and its range

Every quadratic function has a domain of all real numbers, so the domain of \(f(x) = -4x^{2} + 8x - 3\) is \(-\infty < x < \infty\). Since the function has a maximum value, the range is \(-\infty < f(x) \leq 1\) or \(f(x) \in (-\infty, 1]\).

Key Concepts

Maximum Value of a Quadratic Function

Understanding the maximum value of a quadratic function is essential, as it represents the highest point on the graph of a parabola that opens downward. In the case of the quadratic function f(x) = -4x^2 + 8x - 3, we first observed that the coefficient of the x^2 term is negative (-4), indicating that the parabola indeed opens downward.

To locate the maximum value, we need to find the vertex, which is the highest point when the parabola opens downward. The x-coordinate of this vertex is determined by the formula \( x = -\frac{b}{2a} \). After calculating the x-coordinate as \( x = 1 \), we substitute this back into the original equation to find the maximum value, \( f(1) = 1 \). Therefore, the maximum value of this function is 1, occurring at \( x = 1 \).

Recognizing the maximum value is crucial for a variety of applications, including optimizing area and revenue, or assessing the highest possible outcome in a given scenario.

Vertex of a Parabola

The vertex of a parabola is a point that can represent either the minimum or maximum value of the quadratic function, depending on the direction the parabola opens. For f(x) = -4x^2 + 8x - 3, the vertex is the point that gives us the maximum value since the parabola opens downward.

Using the formula \( x = -\frac{b}{2a} \), we have found that the vertex occurs at \( x = 1 \). To find the corresponding y-coordinate (the maximum value), we substituted \( x = 1 \) into the function, yielding \( f(1) = 1 \). Hence, the vertex of this particular parabola is at the point \( (1, 1) \). The concept of a vertex is integral in graphing quadratic functions and in solving real-world problems involving maximum and minimum values.

Domain and Range of Quadratic Function

The domain and range of a quadratic function are two fundamental concepts that describe the set of allowable x-values (domain) and the resulting y-values (range) that the function can have. In every quadratic function, the domain is all real numbers, which can be denoted as \( -\infty < x < \infty \), indicating that for any real number x, there exists a corresponding y-value.

For the given function f(x) = -4x^2 + 8x - 3, the downward-opening parabola indicates that the y-values have an upper limit, which corresponds to the maximum value of the function. As such, the range of our function is \( -\infty < f(x) \leq 1 \) or, in interval notation, \( f(x) \in (-\infty, 1] \). Understanding the domain and range of a quadratic function is crucial when modeling real-world situations to determine the potential scope of outcomes.