Problem 50 - Algebra 2

Question

Determine whether each function has a maximum or a minimum value and find the maximum or minimum value. Then state the domain and range of the function. $$ f(x)=\frac{3}{4} x^{2}-5 x-2 $$

Step-by-Step Solution

Answer

The function has a minimum value of \(-\frac{31}{3}\). Domain: \((-\infty, \infty)\); Range: \([-\frac{31}{3}, \infty)\).

Step 1: Identify the Type of Function

The given function is a quadratic function of the form \( f(x) = ax^2 + bx + c \) where \( a = \frac{3}{4} \), \( b = -5 \), and \( c = -2 \). Since \( a > 0 \), the parabola opens upwards, indicating that the function has a minimum value.

Step 2: Calculate the Vertex (Minimum Value)

The vertex of a quadratic function \( f(x) = ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \). Substitute \( a = \frac{3}{4} \) and \( b = -5 \) into the formula to find \( x \): \[ x = -\frac{-5}{2 \times \frac{3}{4}} = \frac{5}{\frac{3}{2}} = \frac{10}{3} \]. Now, substitute \( x = \frac{10}{3} \) back into the function to find \( f(x) \):

Step 3: Evaluate the Minimum Value

Substitute \( x = \frac{10}{3} \) into the function: \[ f\left(\frac{10}{3}\right) = \frac{3}{4}\left(\frac{10}{3}\right)^2 - 5\left(\frac{10}{3}\right) - 2 \].1. Calculate \( \left(\frac{10}{3}\right)^2 = \frac{100}{9} \).2. Then \( \frac{3}{4} \times \frac{100}{9} = \frac{300}{36} = \frac{25}{3} \).3. Calculate \( 5 \times \frac{10}{3} = \frac{50}{3} \).4. Substitute back: \( f\left(\frac{10}{3}\right) = \frac{25}{3} - \frac{50}{3} - 2 \).5. Simplify: \( \frac{25}{3} - \frac{50}{3} = -\frac{25}{3} \).6. Complete calculation: \(-\frac{25}{3} - 2 = -\frac{25}{3} - \frac{6}{3} = -\frac{31}{3} \).Thus, the minimum value is \(-\frac{31}{3}\).

Step 4: Determine the Domain and Range

The domain of a quadratic function is all real numbers, since there are no restrictions on \( x \). Therefore, the domain is \((-\infty, \infty)\).The range of the function, since it has a minimum value of \(-\frac{31}{3}\) and opens upwards, is \([-\frac{31}{3}, \infty)\).

Key Concepts

Vertex Calculation

In a quadratic function like \( f(x) = \frac{3}{4}x^2 - 5x - 2 \), finding the vertex is essential. The vertex of a parabola gives us the point where the graph changes direction. If the parabola opens upwards, like in this case (since \( a = \frac{3}{4} > 0 \)), the vertex will give us the minimum point. To find the vertex, use the formula for the \( x \)-coordinate:

• \( x = -\frac{b}{2a} \)

Plugging in \( a = \frac{3}{4} \) and \( b = -5 \), the calculation becomes:

• \( x = -\frac{-5}{2 \times \frac{3}{4}} = \frac{5}{\frac{3}{2}} = \frac{10}{3} \).

Once you've found \( x \), substitute it back into the function to find the \( y \)-coordinate, providing you with the complete vertex \( \left( \frac{10}{3}, f\left(\frac{10}{3}\right) \right) \). The vertex is a turning point and greatly aids in graphing the function.

Domain and Range

Understanding the domain and range of a quadratic function is key in analyzing how the function behaves. The domain of any quadratic function, including \( f(x) = \frac{3}{4}x^2 - 5x - 2 \), is always all real numbers \((-\infty, \infty)\). This is because you can insert any real number into the function for \( x \) without any restrictions. For its range, you must consider the direction in which the parabola opens. Since \( a = \frac{3}{4} > 0 \), the parabola opens upwards. This means its range starts from the minimum value located at the vertex and extends to positive infinity. Given that the minimum value is \(-\frac{31}{3} \), the range is:

• \([-\frac{31}{3}, \infty)\)

Ensuring you understand domain and range helps you predict the behavior of the quadratic function across the graph.

Minimum Value

Identifying the minimum value of a quadratic function helps understand the lowest point on its graph. For the function \( f(x) = \frac{3}{4}x^2 - 5x - 2 \), it happens at the vertex. Determining the vertex's \( y \)-coordinate provides the minimum value. After finding \( x = \frac{10}{3} \) for the vertex using \( x = -\frac{b}{2a} \), substitute this back into the function to find \( f\left(\frac{10}{3}\right) \):

• Calculate \( \left(\frac{10}{3}\right)^2 = \frac{100}{9} \)
• Then \( \frac{3}{4} \times \frac{100}{9} = \frac{25}{3} \)
• Next, \( 5 \times \frac{10}{3} = \frac{50}{3} \)
• Substitution yields \( f\left(\frac{10}{3}\right) = \frac{25}{3} - \frac{50}{3} - 2 \)
• Simplifying gives \(-\frac{31}{3}\)

Thus, the minimum value of the function is \(-\frac{31}{3}\). Recognizing this minimum value is crucial in understanding the quadratic function's characteristics, especially for determining its range and assessing graphical interpretations.