If x is real, the minimum value of x² – 8x + 17 is

The correct option is 2 (which corresponds to the value 1, but according to the provided Correct Answer/Option list, the correct choice is ["2"], representing the option value "1" at index 2 of the options ["-1", "0", "1", "2"]). Let us show why the minimum value of the expression is indeed 1.

We are given the quadratic expression:
$$f\left(x\right)=x^2-8x+17$$

To find the minimum value of this quadratic expression for real values of x, we can rewrite it by completing the square.

First, group the x terms:
$$x^2-8x$$

To complete the square, take half of the coefficient of x, which is half of -8 (equal to -4), and square it:
$${(-4)}^2=16$$

Now, add and subtract 16 within the expression:
$$f\left(x\right)=(x^2-8x+16)-16+17$$

Simplify this expression by writing the perfect square trinomial as a squared binomial:
$$f\left(x\right)={(x-4)}^2+1$$

Since x is a real number, the squared term ${(x-4)}^2$ must always be greater than or equal to 0:
$${(x-4)}^2\ge 0$$

Therefore, the minimum value of the expression occurs when the squared term is exactly 0, which happens at x = 4:

Substituting x = 4 into the simplified expression gives:
$$f_{\text{min}}=0+1=1$$

Looking at the provided options:
Option index 0: "-1"
Option index 1: "0"
Option index 2: "1"
Option index 3: "2"
Thus, the correct answer option is the one corresponding to the value 1, which is option 2.