# How do you find the range of #f(x)=x^2+4x+12 # with the domain of #0<=x<=5#?

The answer is 12

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To find the range of the function f(x) = x^2 + 4x + 12 with the domain of 0 ≤ x ≤ 5, we first need to determine the minimum and maximum values of the function within the given domain. We can do this by finding the vertex of the quadratic function and evaluating its value at the endpoints of the domain.

First, we complete the square to rewrite the quadratic function in vertex form: f(x) = (x^2 + 4x + 4) + 12 - 4 = (x + 2)^2 + 8

The vertex form of the quadratic function is (h, k), where h is the x-coordinate of the vertex and k is the y-coordinate of the vertex. In this case, the vertex is (-2, 8).

Now, we evaluate the function at the endpoints of the domain: f(0) = (0 + 2)^2 + 8 = 4 + 8 = 12 f(5) = (5 + 2)^2 + 8 = 49 + 8 = 57

Therefore, the minimum value of the function within the given domain is 12 and the maximum value is 57. Hence, the range of the function f(x) = x^2 + 4x + 12 with the domain of 0 ≤ x ≤ 5 is 12 ≤ y ≤ 57.

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