84.4k views
0 votes
What are the domain and range of the function f(x)= x^2-4x-12 over x+2

User David Bern
by
7.8k points

1 Answer

7 votes

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
by
7.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories