Questions: Consider the function f(x)=2x^2-16x-9. 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.

Solution Steps

To solve the given problem, we need to analyze the quadratic function \( f(x) = 2x^2 - 16x - 9 \).

a. Determine whether the function has a minimum or maximum value:

• For a quadratic function \( ax^2 + bx + c \), if \( a > 0 \), the parabola opens upwards and has a minimum value. If \( a < 0 \), the parabola opens downwards and has a maximum value. Here, \( a = 2 \), which is greater than 0, so the function has a minimum value.

b. Find the minimum value and determine where it occurs:

• The vertex of the parabola \( f(x) = ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). We can use this to find the x-coordinate of the vertex and then substitute it back into the function to find the minimum value.

c. Identify the function's domain and range:

• The domain of any quadratic function is all real numbers. The range can be determined from the minimum value found in part b.

The function \( f(x) = 2x^2 - 16x - 9 \) is a quadratic function where \( a = 2 \). Since \( a > 0 \), the parabola opens upwards, indicating that the function has a minimum value.

To find the x-coordinate of the vertex, we use the formula: \[ x = -\frac{b}{2a} = -\frac{-16}{2 \cdot 2} = -\frac{16}{4} = -4 \] Next, we substitute \( x = -4 \) back into the function to find the minimum value: \[ f(-4) = 2(-4)^2 - 16(-4) - 9 = 2 \cdot 16 + 64 - 9 = 32 + 64 - 9 = 87 \] Thus, the minimum value is \( 87 \) and it occurs at \( x = -4 \).

The domain of the function is all real numbers, expressed as: \[ \text{Domain} = \mathbb{R} \] The range starts from the minimum value and extends to infinity: \[ \text{Range} = [87, \infty) \]

Final Answer