How do you determine the maximum or minimum value, as well as the domain and range, of the function y = 2x^2 + 32x + 4?

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To identify the maximum or minimum value of the function given by the equation y = 2x2 + 32x + 4, we first need to recognize that this is a quadratic function in standard form, y = ax2 + bx + c, where:

  • a = 2
  • b = 32
  • c = 4

Since the coefficient of x2 (i.e., a) is positive (a = 2), the parabola opens upward. This indicates that the function has a minimum value, rather than a maximum.

Finding the Vertex

The vertex of the parabola (where the minimum value occurs) can be found using the formula:

x = -b / (2a)

Plugging in our values:

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

Now, we can substitute this x-value back into the original equation to find the corresponding y-value:

y = 2(-8)2 + 32(-8) + 4

Calculating further:

y = 2(64) – 256 + 4 = 128 – 256 + 4 = -124

Thus, the minimum value of the function is -124, occurring at x = -8.

Domain and Range

Domain:

A quadratic function is defined for all real numbers, hence the domain of the function is:

Domain: (-∞, ∞)

Range:

Since the parabola opens upward and has a minimum value of -124, the range of the function is:

Range: [-124, ∞)

In summary, for the function y = 2x2 + 32x + 4:

  • Minimum Value: -124 at x = -8
  • Domain: (-∞, ∞)
  • Range: [-124, ∞)

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