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 6 - 2019 - Paper 1

To sketch the curve:

1. Identify the vertex, derived from the completed square form $f(x) = 2(x + 1)^2 + 7$, which is at the point (-1, 7).
2. Find the y-intercept by evaluating f(0): $f(0) = 2(0)^2 + 4(0) + 9 = 9.$ Hence, the y-intercept is at (0, 9).
3. The curve opens upward as the coefficient of the squared term is positive.
4. Next, check the x-intercepts by setting f(x) = 0: $2x^2 + 4x + 9 = 0,$ but since the discriminant (b² - 4ac) is negative, there are no real roots.
5. Plot these points, marking the vertex (-1, 7) and y-intercept (0, 9), ensuring the curve is U-shaped.

Given the function from part c: $g(x) = 2(x - 2)^2 - 4x - 3$, we can express the transformation mapping from f(x) to g(x):

1. The function g(x) can be simplified to confirm its relation to f(x).
2. Notably, the mapping involves translations and stretches:
   • A horizontal translation to the right by 2 units.
   • A vertical translation downwards by adjusting the function accordingly where necessary.

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

1. First, identify the minimum value of the denominator. From the function $f(x) = 2x^2 + 4x + 9$, since it opens upwards, calculate its vertex: $x = -\frac{b}{2a} = -\frac{4}{2(2)} = -1.$
2. Substituting back to find the minimum: $f(-1) = 2(-1)^2 + 4(-1) + 9 = 2 - 4 + 9 = 7.$
3. Therefore, the function approaches 0 but never reaches it, while the maximum value occurs when the denominator is at its minimum.
4. Thus, the range is: $0 < h(x) \leq \frac{21}{7} = 3,$
   or in interval notation:
   $(0, 3].$