MATHEMATICS

QUESTION:

Graphs

Solve the quadratic equation

4 x2 + 4 x – 15 = 0

Graphically given that two solutions lie in the range x = -3 to x = 2. Determine also the co-

ordinates and nature of the turning point of the curve.

Solve graphically the equations

1.20x + y = 1.80

x – 0.50y = 8.50

ANSWER:

x=−b±√b2−4 ac2a

Find coefficients a, b and c.

 a=4, b=4, c=−15

Plug in the values for a, b, and c into the quadratic formula

x1, 2 = − b ±√ b 2−4 ac

2a

x1, 2 = −4±√42−4x4(−15)

2x4

Simplify expression under the square root.

x1, 2 = −4±√256

8

Solve for x

x1 =−4   +   √256

8 = 32

= 1.5

x2 = −4   −   √256

8 = - 52

= -2.5

THE CO-ORDINATES:

4 x2 + 4 x – 15 = 0

4 (-3)2 + 4 (-3) – 15

4 (9) - 12 -15

36 - 12 - 15

36- 27

y = 9

4 x2 + 4 x – 15 = 0

4 (-2)2 + 4 (-2) – 15

4 (4) - 8 -15

16 – 23

y = -7

4 x2 + 4 x – 15 = 0

4 (-1)2 + 4 (-1) – 15

4 (1) - 4 -15

4-19

y = -15

4 x2 + 4 x – 15 = 0

4 (0)2 + 4 (0) – 15

0 + 0 - 15

y = -15

4 x2 + 4 x – 15 = 0

4 (1)2 + 4 (1) – 15

4 +4 -15

8 – 23

y = -7

4 x2 + 4 x – 15 = 0

4 (2)2 + 4 (2) – 15

4 (4) - 8-15

16 + 8 -15

y = 9

x -3 -2 -1 0 1 2

y 9 -7 -15 -15 -7 9

NATURE OF THE TURNING POINT OF THE CURVE:

−b2 a = −(4)

2(4) = −48 = - 1

2

GRAPH:

x -3 -2 -1 0 1 2

y 9 -7 -15 -15 -7 9

-4 -3 -2 -1 0 1 2 3

-20

-15

-10

-5

0

5

10

15

SOLVE GRAPHICALLY:

1.20x + y = 1.80 (1)

x – 0.50y = 8.50 (2) Multiply x 2

1.20x + y = 1.80 (1)

2x – y = 17 (3)

Substitute for y in equation

1.20x – 17 = 2x = 1.80

1.20x + 2x = 1.80 + 17

3.2x = 18.80

x = 18.80/3.2

x= 5.875

Substitute for x in equation

y = -17 + 2 (5.875)

y = -17 + 11.75

y = -5.25

TURNING POINT:

−b2 a = −(5.25)

2(5.88) = −5.2511.76 = - 0.45

GRAPH:

x 5.88

y -5.25

5 6 7 8 9 10 11 12

-6

-5

-4

-3

-2

-1

0