Correct answer: D. f(x) = (x+3)^2 - 4
The question asks for an equivalent form of the function f(x) = x^2 + 6x + 5 that displays the minimum value of the function. The form that shows the minimum or maximum value of a quadratic function is the vertex form, f(x) = a(x-h)^2 + k, where k is the minimum or maximum value. To convert the given function into vertex form, one can use the method of completing the square.
1. Start with the given function: f(x) = x^2 + 6x + 5.
2. Take half of the coefficient of the x-term (which is 6), and square it: (6/2)^2 = 3^2 = 9.
3. Add and subtract this value to the expression to create a perfect square trinomial without changing the value of the expression: f(x) = (x^2 + 6x + 9) - 9 + 5.
4. Factor the perfect square trinomial and combine the constant terms: f(x) = (x+3)^2 - 4.
This is the vertex form of the function. The minimum value of the function is -4, which is displayed as a constant in this form. Therefore, f(x) = (x+3)^2 - 4 is the correct choice.
A is incorrect. This form results from making a sign error in the squared binomial. The expansion of (x-3)^2 is x^2 - 6x + 9, which does not match the original function's x-term.
B is incorrect. This form results from a common error in completing the square, where 9 is added to create the perfect square trinomial, but the corresponding 9 is not subtracted from the constant term, leading to an inequivalent expression.
C is incorrect. This is the factored or intercept form of the function. While it is an equivalent expression to the original function, it displays the x-intercepts (at x=-1 and x=-5) rather than the minimum value of the function.