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# S.4 Mathematics - PowerPoint PPT Presentation

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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### S.4 Mathematics

(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

x + y –7 = 0

x

0

2x – 3y + 6 = 0

(3, 4)

What is the solution of the

simultaneous equations?

x

0

Two points of intersection

x

0

One point of intersection

0

No points of intersection

x

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

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

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

Parabola:

Straight line:

There are two points of intersection

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

Parabola:

Straight line:

# 1 parabola and the straight line.

There is no point of intersection

# 2 parabola and the straight line.

There are two points of intersection

# 3 parabola and the straight line.

There is one point of intersection

y parabola and the straight line. = – 3 x – 6

No point of intersection

y parabola and the straight line. = 2 x – 4

2

3

– 4

2

Two points of intersection

2 parabola and the straight line.x – y – 12= 0

2

3

4

– 8

– 6

– 4

One point of intersection

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?

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?

#2 parabola and the straight line.

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

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)

Exercise points of intersection of the parabola and the straight line?

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

# Ex.1 points of intersection of the parabola and the straight line?

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

24 – 4k = 0

∴ k = 6

Exercise points of intersection of the parabola and the straight line?

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

# Ex.2 points of intersection of the parabola and the straight line?

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

28 + 4k < 0

∴ k < – 7

Exercise points of intersection of the parabola and the straight line?

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.

# Ex.3 points of intersection of the parabola and the straight line?

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