Power of Maths Paper 1 OL

Quadratic Functions

17 EXAMPLE 8

Find the co-ordinates of the point V which gives y = 3x 2 – 12x + 5 its minimum value.

Solution y = 3x 2 – 12x + 5

a = 3, b = –12, c = 5 Vertex V

As a > 0, the quadratic graph is cup-shaped and so the vertex is the point at which the minimum value of the function occurs. The minimum value of the function is the value of the y co-ordinate at this point.

At the vertex: x = − b ___

2a = − (−12) x = 2: y = 3(2)2 – 12(2) + 5 = –7

Therefore, the co-ordinates of the vertex are V (2, –7).

The minimum value of the function is –7.

______ 6

= 2 EXAMPLE 9

Find the co-ordinates of the point V which gives y = –2x 2 – x + 7 its maximum value.

Solution y = –2x 2 – x + 7

a = –2, b = –1, c = 7

As a < 0, the quadratic graph is cap-shaped and so the vertex is the point at which the maximum value of the function occurs. The maximum value of the function is the value of the y co-ordinate at this point.

At the vertex: x = – b ___

x = − 1 __

The maximum value of the function is 57 ___

( − 1 __

4 , 57 ___

8 ) . EXAMPLE 10

A ball is projected vertically upwards from the top of a 13 m high cliff. The ball’s height h, in metres above the ground, t seconds after it is launched is given by: h = –10t 2 + 20t + 13

(a) Find its maximum height above the ground.

(b) Find when it hits the ground, correct to two decimal places.

Solution h = –10t 2 + 20t + 13

V

a = –10, b = 20, c = 13 As a = –10 < 0, the quadratic graph is cap-shaped. (a) The maximum height occurs when:

t = − b ___

2a = −20 ______

2(−10) = 1 t = 1: h = (–10)(1)2 + 20(1) + 13 = 23 m

(b) h = 0: –10t 2 + 20t + 13 = 0 10t 2 – 20t – 13 = 0

t =

_________________________ 20

20 ± √

____________________ (−20) 2 − 4 × 10 × (−13)

= 2·52 or – 0·52 [Reject the negative value.] ∴ t = 2·52 s

It hits the ground after 2·52 s. 4 : y = –2

( − 1 __

2a =

4 ) − 2

_____ 2(−2) = − 1

– (–1)

( − 1 __

4

__ 4

) + 7 = 57 ___

8

Therefore, the co-ordinates of the vertex are V

8 . Vertex V

259

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