Given the function: f(x) = 2x^2 + 4x + 9 (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 curve y = f(x) is a quadratic function with a minimum point.

1. Identifying the Vertex: The vertex occurs at x = -1, where the minimum value is f(-1) = 7.
2. Intercepts: - To find the y-intercept, set x = 0:
   $f(0) = 2(0)^2 + 4(0) + 9 = 9$
   • To find x-intercepts, solve:
     $2x^2 + 4x + 9 = 0$
     Using the discriminant:
     $D = b^2 - 4ac = 4^2 - 4(2)(9) = 16 - 72 = -56$ (no real solutions).
3. Sketch: The graph is a U-shaped curve opening upwards, with the vertex at (-1, 7) and the y-intercept at (0, 9).

To analyze the transformation from y = f(x) to y = g(x), we observe that g(x) = 2(x - 2)² - 4x - 3.

1. Translation: It includes a horizontal shift right by 2 units, thereby translating the graph of f(x).
2. Vertical Stretch and Reflection: The '2' in front of (x - 2)² indicates a vertical stretch of 2.
3. Vertical Shift: The additional terms -4x - 3 suggest a further vertical transformation but need careful adjustment.

To find the range of h(x) = \frac{21}{2x^2 + 4x + 9}, we first note that the denominator 2x² + 4x + 9 is positive for all x (as it has no real roots).

1. Minimum Value of Denominator: The minimum occurs at x = -1:
   $2(-1)^2 + 4(-1) + 9 = 2 - 4 + 9 = 7$
2. Maximum Value of h(x): The maximum value occurs when the denominator is at its minimum:
   $h_{max} = \frac{21}{7} = 3$
3. Therefore, as x approaches ±∞, h(x) approaches 0. Thus, the range is:
   $0 < h(x) ext{ (maximum value 3)}$