Find the minimum value of the function f(x)=x^(2)+12 x+38 to the nearest hundredth. Answer:

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Find the minimum value of the function  f(x)=x^(2)+12 x+38 to the nearest hundredth. Answer:

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Complete the Square: To find the minimum value of the quadratic function f(x) = x^2 + 12x + 38, we can complete the square to rewrite the function in vertex form, which will give us the vertex of the parabola. The vertex form of a quadratic function is f(x) = a(x - h)^2 + k, where (h, k) is the vertex of the parabola. Since the coefficient of x^2 is 1 (a = 1), the function is already in the correct form to complete the square.Calculate Half and Square: First, we take the coefficient of x, which is 12, divide it by 2, and square the result to complete the square. This gives us (12/2)^2 = 6^2 = 36.Add and Subtract: Next, we add and subtract this number inside the function to complete the square. This does not change the function because we are adding and subtracting the same value. f(x) = x^2 + 12x + 36 + 38 - 36 f(x) = (x^2 + 12x + 36) + 2 f(x) = (x + 6)^2 + 2Convert to Vertex Form: Now the function is in vertex form, f(x) = (x + 6)^2 + 2. The vertex of the parabola is at (-6, 2). Since the coefficient of the squared term is positive, the parabola opens upwards, and the vertex represents the minimum point of the function.Find Minimum Value: The minimum value of the function is the y-coordinate of the vertex, which is 2. Since the question asks for the answer to the nearest hundredth, we do not need to round the value because it is already an integer.

Find the minimum value of the function $f(x)=x^{2}+12 x+38$ to the nearest hundredth. $\newline$ Answer:

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Find the minimum value of the function f(x)=x2+12x+38 f(x)=x^{2}+12 x+38 f(x)=x2+12x+38 to the nearest hundredth.\newlineAnswer:

Find the minimum value of the function $f(x)=x^{2}+12 x+38$ to the nearest hundredth. $\newline$ Answer: