Cambridge International AS and A Level Mathematics Pure Mathematics 1

The graphs of quadratic functions

23

P1^

1

Rewrite the expression with the constant term moved to the right

x^2 − 4 x + 5.

Take the coefficient of x: − 4

Divide it by 2: − 2

Square the answer: + 4

Add this to the left-hand part and compensate by subtracting it from the constant

term on the right

x^2 – 4x + 4 + 5 – 4.

This can now be written as (x − 2)^2 + 1.

EXAMPLE 1.31 Write x^2 + 5 x + 4 in completed square form.

Hence state the equation of the line of symmetry and the co-ordinates of the

vertex of the curve y = x^2 + 5 x + 4.

SOLUTION

x^2 + 5 x + 4

x^2 + 5 x + 6.25 + 4 − 6.25

(x + 2.5)^2 − 2.25 (This is the completed square form.)

The line of symmetry is x + 2.5 = 0, or x = −2.5.

The vertex is (−2.5, −2.25).

This is the completed

square form.

The minimum value is 1,

The line of symmetry is so the vertex is (2, 1).

x – 2 = 0 or x = 2.

5 ÷ 2 = 2.5; 2.5^2 = 6.25

± ± ± ± ±    x

±

±

±



x ± 

y

Figure 1.8

Vertex

(–2.5, –2.25)

Line of symmetry

x = –2.5