Past Exam Questions

To express \(x^2 - 4x + 8\) in the form \((x-a)^2 + b\), we complete the square:

1. Start with the expression: \(x^2 - 4x + 8\).

2. Take the coefficient of \(x\), which is \(-4\), halve it to get \(-2\), and square it to get \(4\).

3. Add and subtract \(4\) inside the expression:

\(x^2 - 4x + 4 - 4 + 8\)

4. Rewrite as a perfect square and simplify:

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

To express \(3x^2 - 12x + 7\) in the form \(a(x + b)^2 + c\), follow these steps:

1. Factor out the coefficient of \(x^2\) from the quadratic and linear terms:

\(3(x^2 - 4x) + 7\)

2. Complete the square for the expression inside the parentheses:

\(x^2 - 4x = (x - 2)^2 - 4\)

3. Substitute back into the expression:

\(3((x - 2)^2 - 4) + 7\)

4. Distribute the 3 and simplify:

\(3(x - 2)^2 - 12 + 7\)

5. Combine the constants:

\(3(x - 2)^2 - 5\)

Thus, \(a = 3\), \(b = -2\), and \(c = -5\).

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

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

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

3. Add and subtract this square inside the expression: \(x^2 + 6x + 9 - 9 + 2\).

4. Rewrite as a perfect square: \((x + 3)^2 - 9 + 2\).

5. Simplify the constants: \((x + 3)^2 - 7\).

Thus, \(a = 3\) and \(b = -7\).

Given the function \(f(x) = x^2 + ax + b\) with roots \(x = 1\) and \(x = 9\), we can express it as:

\(f(x) = (x - 1)(x - 9)\)

Expanding this, we get:

\(f(x) = x^2 - 10x + 9\)

Thus, \(a = -10\) and \(b = 9\).

To find the vertex, use the vertex formula \(x = \frac{-b}{2a}\) or symmetry:

\(x = \frac{1 + 9}{2} = 5\)

Substitute \(x = 5\) into \(f(x)\):

\(f(5) = 5^2 - 10 \times 5 + 9 = -16\)

Thus, the vertex is \((5, -16)\).

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.

To express \(x^2 - 4x + 5\) in the form \((x + a)^2 + b\), we complete the square:

Start with the expression:

\(x^2 - 4x + 5\)

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

\(x^2 - 4x + 4 - 4 + 5\)

Rewrite as a perfect square:

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

Thus, the expression in the form \((x + a)^2 + b\) is \((x - 2)^2 + 1\).

The minimum point occurs at \(x = 2\), giving the coordinates \((2, 1)\).

To express \(2x^2 - 4x + 1\) in the form \(a(x + b)^2 + c\), we complete the square:

Start with the quadratic expression:

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

Factor out the 2 from the first two terms:

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

Complete the square inside the parentheses:

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

Substitute back:

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

Distribute the 2:

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

Simplify:

\(2(x - 1)^2 - 1\)

Thus, \(a = 2\), \(b = -1\), \(c = -1\).

The minimum point \(A\) is at \((1, -1)\).