To express \(2x^2 - 8x + 14\) in the form \(2[(x-a)^2 + b]\), we follow these steps:

1. Start with the expression: \(2x^2 - 8x + 14\).

2. Factor out the 2 from the first two terms: \(2(x^2 - 4x) + 14\).

3. Complete the square inside the parentheses:

- Take half of the coefficient of \(x\), which is \(-4\), giving \(-2\).

- Square it to get \(4\).

4. Add and subtract this square inside the parentheses:

\(2[(x^2 - 4x + 4) - 4] + 14\).

5. Simplify the expression:

\(2[(x-2)^2 - 4] + 14\).

6. Distribute the 2 and simplify:

\(2(x-2)^2 - 8 + 14\).

7. Combine the constants:

\(2(x-2)^2 + 6\).

8. Rewrite in the desired form:

\(2[(x-2)^2 + 3]\).

To express \(5y^2 - 30y + 50\) in the form \(5(y + a)^2 + b\), follow these steps:

1. Start with the expression: \(5y^2 - 30y + 50\).

2. Factor out 5 from the first two terms: \(5(y^2 - 6y) + 50\).

3. Complete the square inside the parentheses:

- Take half of the coefficient of \(y\), which is \(-6\), to get \(-3\).

- Square \(-3\) to get 9.

4. Add and subtract 9 inside the parentheses:

\(5(y^2 - 6y + 9 - 9) + 50\).

5. Rewrite the expression as:

\(5((y - 3)^2 - 9) + 50\).

6. Distribute the 5:

\(5(y - 3)^2 - 45 + 50\).

7. Simplify the constants:

\(5(y - 3)^2 + 5\).

Thus, \(a = -3\) and \(b = 5\).

To express \(16x^2 - 24x + 10\) in the form \((4x + a)^2 + b\), follow these steps:

1.

Start by expanding \((4x + a)^2\):

\((4x + a)^2 = 16x^2 + 8ax + a^2\) 2.

Compare coefficients with \(16x^2 - 24x + 10\):

- The coefficient of \(x\) gives: \(8a = -24\) - Solve for \(a\):

\(a = \frac{-24}{8} = -3\) 3.

Substitute \(a = -3\) back into \((4x + a)^2\):

\((4x - 3)^2 = 16x^2 - 24x + 9\) 4.

Find \(b\) by comparing the constant terms:

\(16x^2 - 24x + 10 = (4x - 3)^2 + b\) - From \(9 + b = 10\), solve for \(b\):

\(b = 1\) 5.

Thus, the expression is:

\((4x - 3)^2 + 1\)

To express \(x^2 + 6x + 5\) in the form \((x + a)^2 + b\), we complete the square:

Start with the expression: \(x^2 + 6x + 5\).

1. Take half of the coefficient of \(x\), which is 6, giving 3, and square it: \(3^2 = 9\).

2. Add and subtract this square inside the expression:

\(x^2 + 6x + 9 - 9 + 5\).

3. Rewrite the expression as a perfect square and a constant:

\((x + 3)^2 - 4\).

Thus, \(a = 3\) and \(b = -4\).

Given the equation \(y = 2x^2 + kx + k - 1\) and \(k = 2\), substitute \(k\) to get:

\(y = 2x^2 + 2x + 2 - 1 = 2x^2 + 2x + 1\)

We need to express this in the form \(y = 2(x + a)^2 + b\).

Start by completing the square for \(2x^2 + 2x\):

\(2(x^2 + x) = 2\left((x + \frac{1}{2})^2 - \frac{1}{4}\right)\)

\(= 2(x + \frac{1}{2})^2 - \frac{1}{2}\)

Now, add the constant term:

\(y = 2(x + \frac{1}{2})^2 - \frac{1}{2} + 1\)

\(= 2(x + \frac{1}{2})^2 + \frac{1}{2}\)

Thus, \(a = \frac{1}{2}\) and \(b = \frac{1}{2}\).

The vertex form \(y = 2(x + a)^2 + b\) gives the vertex at \((-a, b)\).

Therefore, the vertex is \(\left(-\frac{1}{2}, \frac{1}{2}\right)\).