f(x) = 2x^2 + 4x + 9 x ∈ ℝ (a) Write f(x) in the form α(x + b)^2 + c, where a, b and c are integers to be found - Edexcel - A-Level Maths Pure - Question 7 - 2019 - Paper 1

The sketch of the curve involves finding key points and the shape of the function:

1. Identify intercepts:

• Y-intercept: Set x = 0.
• $f(0) = 2(0)^2 + 4(0) + 9 = 9$, thus the y-intercept is (0, 9).

2. Turning points:

• The turning point is given by the vertex of the parabola.
• The vertex is at the point (-1, 7). Since this is the minimum point,
• The coordinates of the turning point are (-1, 7).

3. Shape of the curve:

• The graph is U-shaped, opening upwards, indicating that it has a minimum and no maximum.

To describe the transformation from y = f(x) to y = g(x):

1. Identify g(x):

• The transformed function is given by:
• $g(x) = 2(x - 2)^2 - 4x - 3$

2. Comparison with f(x):

• Horizontal translation: The term (x - 2) indicates a shift to the right by 2 units.
• Vertical transformation: The coefficients and constants modify the output value. In describing this, I note that it also involves a combination of stretching and shifting both vertically and horizontally.

To find the range of the function
$h(x) = \frac{21}{2x^2 + 4x + 9}$:

1. First identify the minimum value of the denominator:
   • The quadratic in the denominator, $2x^2 + 4x + 9,$ is positive for all x. To locate the minimum value:
   • This form completes the square to find the minimum at the vertex:
   • Completing gives the minimum value of the quadratic as 3.
2. Identify h(x)'s maximum:
   • The maximum value of h(x) occurs when the minimum of the denominator is at its smallest:
   • When the denominator is at its minimum of 3, thus
   • $h(x)_{max} = \frac{21}{3} = 7$.
3. Infer the range:
   • Since h(x) approaches infinity as the denominator approaches zero, we thus find that the feasible range is
   • $0 < h(x) < 7$.