$x^2 - 10x + 27$ 可以表示为 $(x + p)^2 + q$ 的形式。求 $p$ 的值和 $q$ 的值。 答案 $p =$ ___ 答案 $q =$ ___ 由此写出函数 $y = x^2 - 10x + 27$ 图像的最低点（顶点）坐标。 答案 ( , )

1. Expand and compare the coefficient of $x$.

$(x+p)^2+q=x^2+2px+p^2+q$
Compare with
$x^2-10x+27.$
So,
$2p=-10$
$p=-5$

1. Substitute the known value of $p$ and compare the constant term.

Since $p=-5$,
$(x+p)^2+q=(x-5)^2+q=x^2-10x+25+q.$
Compare with $x^2-10x+27$:
$25+q=27$
$q=2.$

1. Identify the minimum point from the completed-square form.

From part (a),
$y=x^2-10x+27=(x-5)^2+2.$
Since $(x-5)^2 \geq 0$, the minimum value of $y$ is $2$, when $x-5=0$, so $x=5$.

Minimum point: $(5,2)$