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What are the domain and range of the function f(x)= x^2-4x-12 over x+2

User David Bern
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Explanation:

The function f(x) can be written as:

f(x) = (x^2 - 4x - 12) / (x + 2)

The domain of a function is the set of all values that x can take on. In this case, we need to consider values of x that make the denominator of the fraction nonzero. So the domain of the function is all real numbers except x = -2, since division by zero is undefined.

Therefore, the domain of the function is:

x ∈ ℝ, x ≠ -2

To find the range of the function, we can analyze the behavior of the function as x approaches positive or negative infinity. As x approaches positive or negative infinity, the function approaches the value of x^2 / x = x. So the range of the function is all real numbers except possibly one value that the function cannot take on due to the denominator being zero.

Since the function is a quadratic function, we can use the technique of completing the square to find that the minimum value occurs at x = 2, and has a value of -16. We can also see that the function is symmetric about x = 2. Since the function is decreasing to the left of x = 2 and increasing to the right of x = 2, the range of the function is:

y ∈ ℝ, y ≥ -16 \ y

where the notation y means "the set of all y such that ...".

Substituting x = -2 into the function, we get:

f(-2) = (-2)^2 - 4(-2) - 12 / (-2 + 2) = undefined

So we need to remove the undefined value from the range. Therefore, the range of the function is:

y

User George Netu
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