To analyze the function f(x) = 2x² + 32x + 1, we first convert it into vertex form to find the vertex.
1. **Identifying the Vertex:** The standard form of a quadratic function is ax² + bx + c. Here, a = 2, b = 32, and c = 1. The x-coordinate of the vertex can be found using the formula:
x = -b / (2a)
Plugging in the values:
x = -32 / (2 * 2) = -32 / 4 = -8
Now, we substitute this x-value back into the function to find the y-coordinate:
f(-8) = 2(-8)² + 32(-8) + 1
f(-8) = 2(64) – 256 + 1 = 128 – 256 + 1 = -127
Thus, the vertex of the function is at the point (-8, -127).
2. **Domain:** The domain of a quadratic function is all real numbers since it can take any x-value. Therefore, the domain is:
Domain: (-∞, ∞)
3. **Range:** Since the parabola opens upwards (as a = 2 is positive), the minimum value occurs at the vertex. The range starts from the y-coordinate of the vertex and goes to positive infinity:
Range: [-127, ∞)
In summary, for the function f(x) = 2x² + 32x + 1:
- Vertex: (-8, -127)
- Domain: (-∞, ∞)
- Range: [-127, ∞)