将 $4x^{2}+8x-5$ 表示为 $p(x+q)^{2}+r$ 的形式，其中 $p$、$q$ 和 $r$ 为待求常数。 由此，写出曲线 $y=4x^{2}+8x-5$ 的顶点坐标。

1. Factorise the coefficient of $x^2$ from the first two terms

$4x^{2}+8x-5 = 4(x^{2}+2x)-5$

2. Complete the square inside the bracket

$= 4\left[ (x+1)^{2} - 1^{2} \right] - 5$

3. Expand and simplify to reach the final form

$= 4(x+1)^{2} - 4 - 5 = 4(x+1)^{2} - 9$

1. Identify the turning point from the completed square form found in part (i)

From part (i), $y = 4(x+1)^2 - 9$.
The minimum value of $y$ occurs when $x+1 = 0$, which gives $x = -1$.
The minimum value is $y = -9$.
Therefore, the coordinates of the turning point are $(-1, -9)$.