To express \(2x^2 - 12x + 7\) in the form \(a(x+b)^2 + c\), we start by completing the square:

1. Factor out the coefficient of \(x^2\) from the first two terms:

\(2(x^2 - 6x) + 7\)

2. Complete the square inside the parentheses:

\(x^2 - 6x = (x-3)^2 - 9\)

3. Substitute back into the expression:

\(2((x-3)^2 - 9) + 7\)

4. Distribute the 2 and simplify:

\(2(x-3)^2 - 18 + 7\)

\(2(x-3)^2 - 11\)

Thus, \(a = 2\), \(b = -3\), and \(c = -11\).

To express \(2x^2 - 10x + 8\) in the form \(a(x + b)^2 + c\), we complete the square:

Start with the quadratic expression:

\(2x^2 - 10x + 8\)

Factor out the 2 from the first two terms:

\(2(x^2 - 5x) + 8\)

Complete the square inside the parentheses:

\(x^2 - 5x = (x - \frac{5}{2})^2 - (\frac{5}{2})^2\)

\(= (x - \frac{5}{2})^2 - \frac{25}{4}\)

Substitute back into the expression:

\(2((x - \frac{5}{2})^2 - \frac{25}{4}) + 8\)

\(= 2(x - \frac{5}{2})^2 - \frac{25}{2} + 8\)

\(= 2(x - \frac{5}{2})^2 - \frac{25}{2} + \frac{16}{2}\)

\(= 2(x - \frac{5}{2})^2 - \frac{9}{2}\)

Thus, \(a = 2\), \(b = -\frac{5}{2}\), \(c = -\frac{9}{2}\).

The minimum value of \(2x^2 - 10x + 8\) occurs at the vertex of the parabola, which is \(-\frac{9}{2}\).

To express \(4x^2 - 12x\) in the form \((2x + a)^2 + b\), we start by expanding \((2x + a)^2\):

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

We want this to match \(4x^2 - 12x\), so we equate the coefficients:

\(4a = -12\) which gives \(a = -3\).

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

\((2x - 3)^2 = 4x^2 - 12x + 9\).

Now, compare \(4x^2 - 12x + 9\) with \(4x^2 - 12x\):

\(4x^2 - 12x = (2x - 3)^2 - 9\).

Thus, \(b = -9\).

To express \(2x^2 - 3x\) in the form \(a(x + b)^2 + c\), we complete the square:

Start with the equation:

\(y = 2x^2 - 3x\)

Factor out the 2 from the quadratic terms:

\(y = 2(x^2 - \frac{3}{2}x)\)

Complete the square inside the parentheses:

Take half of the coefficient of \(x\), square it, and add and subtract inside the bracket:

\(x^2 - \frac{3}{2}x = \left(x - \frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2\)

Substitute back into the equation:

\(y = 2\left(\left(x - \frac{3}{4}\right)^2 - \frac{9}{16}\right)\)

Distribute the 2:

\(y = 2\left(x - \frac{3}{4}\right)^2 - \frac{9}{8}\)

Thus, the equation in the form \(a(x + b)^2 + c\) is:

\(y = 2\left(x - \frac{3}{4}\right)^2 - \frac{9}{8}\)

The vertex form \(a(x - h)^2 + k\) gives the vertex \((h, k)\).

So, the vertex is \(\left(\frac{3}{4}, -\frac{9}{8}\right)\).

To find the stationary point, we first differentiate \(f(x)\) with respect to \(x\):

\(f'(x) = \frac{d}{dx}[8 - (x - 2)^2] = -2(x - 2)\)

Set the derivative equal to zero to find the stationary point:

\(-2(x - 2) = 0\)

Solving for \(x\), we get:

\(x = 2\)

Substitute \(x = 2\) back into the original function to find the \(y\)-coordinate:

\(f(2) = 8 - (2 - 2)^2 = 8\)

Thus, the coordinates of the stationary point are \((2, 8)\).

To determine the nature of the stationary point, consider the second derivative:

\(f''(x) = \frac{d}{dx}[-2(x - 2)] = -2\)

Since \(f''(x) < 0\), the stationary point at \((2, 8)\) is a maximum.