Express $9 - 8x + x^2$ in the form $a + (x + b)^2$. Find the value of $a$ and of $b$. Answer a = ... b = ... Explain why when $x = 4$, the expression $9 - 8x + x^2$ has its minimum value. Answer ...

1. Rewrite the expression in standard descending powers of $x$.

$9-8x+x^2=x^2-8x+9$

2. Expand the required form and compare coefficients.

$a+(x+b)^2=a+x^2+2bx+b^2$

$=x^2+2bx+(a+b^2)$

Compare with $x^2-8x+9$:

$2b=-8$

3. Find $b$, then use the constant term to find $a$.

$2b=-8$

$b=-4$

$a+b^2=9$

$a+(-4)^2=9$

$a+16=9$

$a=-7$

1. Use the completed-square form to justify the minimum value.

From part (a),
$9-8x+x^2=(x-4)^2-7.$
Since $(x-4)^2\geq 0$ for all real $x$, the smallest possible value of $(x-4)^2$ is $0$. This occurs when $x-4=0$, so $x=4$. Hence the expression has its minimum value when $x=4$.