Cambridge International AS and A Level Mathematics Pure Mathematics 1

Co-ordinate geometry

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EXAMPLE 2.13 Sketch the curve y = x^3 − 3 x^2 − x + 3 = (x + 1) (x − 1) (x − 3).

SOLUTION

Since the polynomial is of order 3, the curve has up to two stationary points. The

term in x^3 has a positive coefficient (+1) and 3 is an odd number, so the general

shape is as shown on the left of figure 2.29.

The actual equation

y = x^3 − 3 x^2 − x + 3 = (x + 1) (x − 1) (x −3)

tells you that the curve:

− crosses the y axis at (0, 3)

− crosses the x axis at (−1, 0), (1, 0) and (3, 0).

This is enough information to sketch the curve (see the right of figure 2.29).

x

y

y x^3 ± 3x^2 + x + 3

±2 ±1 0 1 2 3 4

3

Figure 2.29

n even

coefficient of

xn positive

n odd

coefficient of

xn negative

Figure 2.28