Answer:
Answer B: min. value at -6
Step-by-step explanation:
Complete the square of f(x) = x^2 - 2x - 5.
Note: Please use " ^ " to denote exponentiation, as shown.
Start with f(x) = x^2 - 2x - 5.
Identify the coefficient of the x term: it is 2.
Take half of that and square your result: (1/2)(2) = 1, and then 1^2 = 1.
Add and subtract this 1 between the 2x term and the constant term:
f(x) = x^2 - 2x + 1 - 1 - 5
Rewrite x^2 - 2x + 1 as a perfect square:
f(x) = (x - 1)^2 - 1 - 5, or f(x) = (x - 1)^2 - 6
Compare this to the standard form
f(x) = (x - h)^2 + k
We see that h = 1 and k = -6.
The vertex is located at (h, k); here, it's located at (1, -6).
Thus, the minimum value of this function is at the vertex (1, -6).
This agrees with Answer B: min. value at -6.
