What are the vertex, domain, and range of the function f(x) = 2x^2 + 32x + 1?

To analyze the function f(x) = 2x² + 32x + 1, we’ll identify its vertex, domain, and range.

Finding the Vertex

The vertex of a quadratic function in the form f(x) = ax² + bx + c can be found using the formula:

x = -b / (2a)

For our specific function, we have:

  • a = 2
  • b = 32
  • c = 1

Plugging in the values, we get:

x = -32 / (2 * 2) = -32 / 4 = -8

Now, we can find the corresponding y-coordinate of the vertex by substituting x = -8 back into the function:

f(-8) = 2(-8)² + 32(-8) + 1

f(-8) = 2(64) – 256 + 1 = 128 – 256 + 1 = -127

Thus, the vertex is located at the point (-8, -127).

Finding the Domain

The domain of a quadratic function is all real numbers. Therefore, for the function f(x) = 2x² + 32x + 1, the domain is:

Domain: (-∞, +∞)

Finding the Range

Since the coefficient of (which is 2) is positive, the parabola opens upwards. This means that the vertex represents the minimum point of the function, and from there, it extends infinitely upwards.

Thus, the range is:

Range: [-127, +∞)

Summary

  • Vertex: (-8, -127)
  • Domain: (-∞, +∞)
  • Range: [-127, +∞)

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