Q.If $x$ is real, the minimum value of $x^2 - 8x + 17$ is:
(A) $-1$
(B) $0$
(C) $1$
(D) $2$

Concept: Quadratic function optimization — for $f(x) = ax^2 + bx + c$ with $a > 0$, the minimum occurs at $x = -\frac{b}{2a}$.

Step 1: Identify coefficients: $a = 1$, $b = -8$, $c = 17$.

Step 2: Vertex $x$-coordinate:

$x = -\frac{-8}{2 \cdot 1} = 4$. …

The expression $x^2 - 8x + 17$ is a quadratic in $x$ with a positive coefficient on $x^2$. That means its graph is a parabola opening upward — it has a single lowest point, and no highest point. The question asks for that lowest value.

Why not just differentiate? You could, but completing the square is faster and gives the minimum directly without calculus. It also reveals the expression as a perfect square plus a constant, which is the cleanest way to see the minimum.

Here’s the step-by-step:

1. Complete the square. Take $x^2 - 8x$ and add and subtract the square of half the coefficient of $x$: Half of $-8$ is $-4$, and $(-4)^2 = 16$. So:

2. Interpret the result.

   The term $(x-4)^2$ is always $\ge 0$ for real $x$, and it equals $0$ exactly when $x = 4$.

   Therefore the whole expression is at least $0 + 1 = 1$, and this value is actually attained at $x = 4$.

3. Check the options. …

Method: Finding the Extreme Value of a Quadratic Expression

Use this whenever you need the maximum or minimum value of a quadratic $ax^2+bx+c$, without necessarily needing calculus.

Steps

Step 1: Identify the coefficients $a$, $b$, $c$ and the sign of $a$.

The sign of $a$ tells you immediately whether the parabola opens upward (minimum exists, $a>0$) or downward (maximum exists, $a<0$).

Step 2: Complete the square, or use the vertex formula directly.

Either rewrite the expression as $a\left(x+\tfrac{b}{2a}\right)^2 + \left(c-\tfrac{b^2}{4a}\right)$, or jump straight to the vertex $x$-coordinate:

Step 3: Substitute back to get the extreme value. …

Common Mistakes

Mistake 1: Forgetting to adjust the constant term when completing the square.

Why it's wrong: adding the square of half the $x$-coefficient inside the expression changes its value unless you also subtract the same amount outside — skipping this gives a wrong constant in the final answer. Correct approach: always write it as "add and subtract" the same number, e.g. $x^2-8x+17=(x^2-8x+16)+(17-16)$, keeping the expression equal to the original.

Mistake 2: Reporting the vertex's $x$-value instead of the function's value there. …