Problem 10 - 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. $$ f(x)=-x^{2}-4 x+1 $$

Step-by-Step Solution

Answer

The function has a maximum value of -11, with domain (-∞, ∞) and range (-∞, -11].

Step 1: Identify the Type of Quadratic Function

The function is given as \( f(x) = -x^2 - 4x + 1 \). This is a quadratic function of the form \( ax^2 + bx + c \). Here, \( a = -1 \), \( b = -4 \), and \( c = 1 \).

Step 2: Determine Maximum or Minimum

Since \( a = -1 \) is less than zero, the parabola opens downward. Therefore, the function has a maximum value (not a minimum).

Step 3: Find the Vertex

The vertex of a parabola given by \( f(x) = ax^2 + bx + c \) can be found at \( x = -\frac{b}{2a} \). Plug in \( b = -4 \) and \( a = -1 \) to get \( x = -\frac{-4}{2(-1)} = 2 \).

Step 4: Calculate Maximum Value

Substitute \( x = 2 \) back into the original function to find the maximum value: \( f(2) = -2^2 - 4(2) + 1 = -4 - 8 + 1 = -11 \). Thus, the maximum value is \( -11 \).

Step 5: State the Domain

The domain of any quadratic function is all real numbers. Therefore, the domain of \( f(x) = -x^2 - 4x + 1 \) is \( (-\infty, +\infty) \).

Step 6: State the Range

Since the function has a maximum value at \( y = -11 \), the range of \( f(x) \) is \(( -\infty, -11 ]\).

Key Concepts

Maximum and Minimum Values

When working with quadratic functions, one of the important aspects to determine is whether the function has a maximum or minimum value. This is influenced by the direction in which the parabola opens.

• If the quadratic function's formula, represented as \( ax^2 + bx + c \), has \( a < 0 \), the parabola opens downward. Thus, the function attains a maximum value at its vertex.
• Conversely, if \( a > 0 \), the parabola opens upward, and the function has a minimum value at its vertex.

For the function \( f(x) = -x^2 - 4x + 1 \), we observe that \( a = -1 \). Because \( a \) is negative, the parabola opens downward, indicating that \( f(x) \) has a maximum value.

Domain and Range of Quadratic Functions

Understanding the domain and range of a quadratic function helps in comprehending its behavior over the real number line.

• The domain of any quadratic function is always all real numbers. This is because you can plug in any real number for \( x \) and calculate \( f(x) \). So, for our function \( f(x) = -x^2 - 4x + 1 \), its domain is \((-\infty, +\infty)\).
• On the other hand, the range is determined by whether the function has a maximum or minimum value. Since our function has a maximum value, its range is constrained by this maximum. Without surpassing it, \( f(x) \) can take any value that is less than or equal to its maximum value. Therefore, the range is \( (-\infty, -11] \).

Vertex of a Parabola

The vertex of a parabola is a key feature that determines its maximum or minimum value, where the direction of the parabola alters.
To find the vertex of a quadratic function in the form \( ax^2 + bx + c \), use the formula:\[ x = -\frac{b}{2a} \]For the function \( f(x) = -x^2 - 4x + 1 \), substitute \( b = -4 \) and \( a = -1 \) into the formula. This will give:\[ x = -\frac{-4}{2(-1)} = 2 \]Now, to find the corresponding \( y \)-value (maximum or minimum), substitute \( x = 2 \) back into the function:\[ f(2) = -2^2 - 4(2) + 1 = -11 \]Thus, the vertex of the parabola is at \((2, -11)\), marking the point where \( f(x) \) reaches its maximum value.