Express $x^{2} - 6x + 18$ in the form $(x - p)^{2} + q$. Hence, sketch the graph of $y = x^{2} - 6x + 18$ on the axes below. Indicate the turning point and any axial intercepts clearly.

1. Complete the square and simplify the remaining constant.

$x^{2}-6x+18=(x-3)^{2}-9+18=(x-3)^{2}+9$

1. Use the completed-square form to identify the turning point and shape of the parabola.

From part (a),
$y=x^2-6x+18=(x-3)^2+9.$
Since the coefficient of $(x-3)^2$ is positive, the graph opens upwards.

Turning point: $(3,9)$.

2. Find and state the axial intercepts, then describe the required sketch.

For the $y$-intercept, put $x=0$:
$y=0^2-6(0)+18=18.$
So the $y$-intercept is $(0,18)$.

For the $x$-intercepts, put $y=0$:
$0=(x-3)^2+9.$
This has no real solution since $(x-3)^2\geq 0$, so $(x-3)^2+9>0$ for all real $x$.

Therefore, the graph lies above the $x$-axis and has no $x$-intercepts.