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S.4 Mathematics. y. (3, 4). x + y –7 = 0. x. 0. 2 x – 3 y +6=0. Put (3,4) into x + y –7 =0. Put (3,4) into 2 x –3 y +6 =0. LHS = (2)3 – 3(4) + 6. LHS = 3+4 – 7. = 0. = 0. = RHS. = RHS. (3,4) is the solution of the equations of x + y –7 =0 and

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slide2

y

(3, 4)

x + y –7 = 0

x

0

2x – 3y +6=0

Put (3,4) into x +y –7 =0

Put (3,4) into 2x –3y +6 =0

LHS = (2)3 – 3(4) + 6

LHS = 3+4 – 7

= 0

= 0

= RHS

= RHS

(3,4) is thesolutionof the equations of x +y –7 =0 and

2x –3y +6 =0

slide3

y

x + y –7 = 0

x

0

2x – 3y + 6 = 0

(3, 4)

What is the solution of the

simultaneous equations?

slide4

y

x

0

Two points of intersection

slide5

y

x

0

One point of intersection

slide6

y

0

No points of intersection

x

What is the relationship between the number of pointsof intersection and the value of discriminant?

slide7

y

y

y

x

x

0

0

0

Case 1:2 points of intersection

∆ > 0

Case 2: 1 point of intersection

∆ = 0

Case 3:No point of intersection

∆ < 0

slide8

Example

Determine the number of points of intersection of the parabola and the straight line.

Parabola:

Straight line:

slide9

No need to solve the eq.

There are two points of intersection

slide10

Determine the number of points of intersection of the parabola and the straight line.

Parabola:

Straight line:

slide11

# 1

There is no point of intersection

slide12

# 2

There are two points of intersection

slide13

# 3

There is one point of intersection

slide14

y = – 3 x – 6

No point of intersection

slide15

y = 2 x – 4

2

3

– 4

2

Two points of intersection

slide16

2 x – y – 12= 0

2

3

4

– 8

– 6

– 4

One point of intersection

slide17

Determine the number of points of intersection of the parabola and the straight line.

Parabola:

Straight line:

We can use :

I) Graphical method

II) Discriminant method

Any other method?

slide18

Determine the number of points of intersection of the parabola and the straight line.

Parabola:

Straight line:

We can use :

I) Graphical method

II) Discriminant method

Any other method?

slide19

#2

∴(3,4) and (-2,-8) are the 2 points of intersection.

slide20

Which method is the fastest in determining the number of points of intersection of the parabola and the straight line?

I) Graphical method

II) Discriminant method

III) Solving the simultaneous equations

(Algebraic method)

slide21

Exercise

1. If the parabola y = – x2 + 2x + 5 and the line y = k intersect at one point, find the value of k.

slide22

# Ex.1

If the parabola and the line intersect at one point , then the discriminant equals to zero.

24 – 4k = 0

∴ k = 6

slide23

Exercise

2. If the straight line y = 3x + kdoes not cut the parabola y = x2 – 3 x + 2 at any point, find the range of values of k.

There is no point of intersection

slide24

# Ex.2

There is no point of intersection so the discriminant is less thanzero.

28 + 4k < 0

∴ k < – 7

slide25

Exercise

3. If the straight line 2x – y – 1 = 0 cuts the parabola y = 3 x2 + 5x + k at two points, where k is an integer. Find the largest value of k.

slide26

# Ex.3

There are two points of intersection so the discriminant is greater thanzero.

∴ k < – 0.25

– 3 – 12 k > 0

The largest value of k is – 1

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