将 $2x^2 - 3x - 11$ 表示为 $a(x - h)^2 + k$ 的形式，其中 $a$、$h$ 和 $k$ 为常数。 由此，写出曲线 $y = 2x^2 - 3x - 11$ 的顶点坐标。

1. Factor out the coefficient of $x^2$ from the $x$ terms

$2x^2 - 3x - 11 = 2\left(x^2 - \frac{3}{2}x\right) - 11$

2. Complete the square for the expression inside the bracket

$= 2\left[ \left(x - \frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2 \right] - 11$
$= 2\left[ \left(x - \frac{3}{4}\right)^2 - \frac{9}{16} \right] - 11$

3. Expand and simplify to reach the required form

$= 2\left(x - \frac{3}{4}\right)^2 - 2\left(\frac{9}{16}\right) - 11$
$= 2\left(x - \frac{3}{4}\right)^2 - \frac{9}{8} - \frac{88}{8}$
$= 2\left(x - \frac{3}{4}\right)^2 - \frac{97}{8}$

4. State the coordinates of the turning point

The turning point of $y = a(x - h)^2 + k$ is $(h, k)$.
Turning point = $\left(\frac{3}{4}, -\frac{97}{8}\right)$