How do you identify the vertex, domain, and range for the function f(x) = x^2 + 4?

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The function given is a quadratic function, typically represented as f(x) = ax^2 + bx + c. In this case, we have:

  • a = 1
  • b = 0
  • c = 4

Finding the Vertex

The vertex of a quadratic function can be determined using the formula for the x-coordinate of the vertex, x = -b / (2a). Substituting the values for a and b:

x = -0 / (2 * 1) = 0

Now, substitute x = 0 back into the function to get the y-coordinate of the vertex:

f(0) = (0)^2 + 4 = 4

Thus, the vertex is located at the point (0, 4).

Finding the Domain

The domain of a quadratic function is all real numbers, since there are no restrictions on the value of x.

In interval notation, the domain is expressed as:

Domain: (-∞, ∞)

Finding the Range

The range of a quadratic function depends on the direction in which it opens. Since the coefficient a is positive (a = 1), the parabola opens upwards. The lowest point on the graph is the y-coordinate of the vertex.

Thus, the range starts from the y-coordinate of the vertex and extends to positive infinity:

Range: [4, ∞)

Summary

To summarize:

  • Vertex: (0, 4)
  • Domain: (-∞, ∞)
  • Range: [4, ∞)

This information provides a complete view of the quadratic function f(x) = x^2 + 4.

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