f(x) = x^4 - 4x - 8 - Edexcel - A-Level Maths Pure - Question 1 - 2007 - Paper 6

To find the turning point, we first find the derivative:

$f'(x) = 4x^3 - 4$

Set the derivative to zero to find critical points:

\Rightarrow x^3 = 1 \ \Rightarrow x = 1$$ Next, we substitute x = 1 back into the original function to find the y-coordinate: $$f(1) = (1)^4 - 4(1) - 8 = 1 - 4 - 8 = −11$$ Thus, the coordinates of the turning point are (1, -11).

To find the constants a, b and c, we expand the given form:

$f(x) = (x-2)(x^2 + ax + bx + c) = x^3 + (a + b - 2)x^2 + (c - 2a)x - 2c$

We know that:

• We want the coefficients to match with those in f(x) = x^4 - 4x - 8. Therefore:
  • Coefficient of x^2: $a + b - 2 = 0$
  • Coefficient of x: $c - 2a = -4$
  • Constant: $-2c = -8 \ \Rightarrow c = 4$

Substituting c = 4 into $c - 2a = -4$ gives: $4 - 2a = -4 \ \Rightarrow 2a = 8 \ \Rightarrow a = 4$

Using $a + b - 2 = 0$ with a = 4: $4 + b - 2 = 0 \ \Rightarrow b = -2$

Thus, the values are: a = 4, b = -2, c = 4.

To sketch the graph of y = f(x), we:

1. Mark the turning point (1, -11).
2. Note the behavior as x approaches positive and negative infinity; the ends will rise as it is a quartic function.
3. Plot the y-intercepts and roots identified previously.
4. Ensure the graph passes through the points and shows the correct curvature.

To sketch y = |f(x)|:

1. Mirror any parts of the graph of y = f(x) that are below the x-axis to above the x-axis.
2. For the part of y = f(x) which is above the x-axis, keep it unchanged.
3. The resulting graph will touch the x-axis at the roots found previously and smoothly connect the sections from above.