The curve C has equation $y = f(x)$, where $f'(x) = (x - 3)(3x + 5)$ Given that the point P (1, 20) lies on C, (a) find $f(x)$, simplifying each term - Edexcel - A-Level Maths Pure - Question 1 - 2018 - Paper 1

We have:

$f(x) = x^3 - 2x^2 - 15x + 36$

To factor further, we can substitute $x - 3$:

$f(x) = (x - 3)(Ax^2 + Bx + C)$

Expanding:

$(x - 3)(Ax^2 + Bx + C) = Ax^3 + Bx^2 + Cx - 3Ax^2 - 3Bx - 3C$

Now, collecting like terms gives:

$f(x) = Ax^3 + (B - 3A)x^2 + (C - 3B)x - 3C$

By matching the coefficients with the expanded form $x^3 - 2x^2 - 15x + 36$, we find:

• $A = 1$
• $B - 3A = -2 \, \Rightarrow B = 1$
• $C - 3B = -15 \, \Rightarrow C = -12$
• $-3C = 36 \, \Rightarrow C = -12 (consistent)$

Thus, it shows:

$f(x) = (x - 3)(x + 4)$

To sketch the graph of $C$, consider:

1. Y-intercept: Set $x = 0$: $f(0) = 36$ Thus, the graph cuts the y-axis at $(0, 36)$.

2. X-intercepts: Solve for $f(x) = 0$: $(x - 3)(x + 4) = 0$ gives the points $(3, 0)$ and $(-4, 0)$. Thus, the graph cuts the x-axis at these coordinates.

The sketch will clearly show continuous curves between these intercepts. Mark the coordinates $(0, 36)$ for y-intercept, and points $(3, 0)$ and $(-4, 0)$ for x-intercepts, ensuring the proper cubic shape with appropriate turning points.