S.4 Mathematics x + y –7 = 0 2x – 3y +6=0 x y 0 (3, 4) Put (3,4) into x +y –7 =0 LHS = 3+4 – 7 Put (3,4) into 2x –3y +6 =0 LHS = (2)3 – 3(4) + 6 = 0.

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Presentation transcript:

S.4 Mathematics

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

x + y –7 = 0 2x – 3y + 6 = 0 x y 0 (3, 4) What is the solution of the simultaneous equations?

x y 0 Two points of intersection

x y 0 One point of intersection

x y 0 What is the relationship between the number of points of intersection and the value of discriminant? No points of intersection

x y 0 x y 0 y 0 Case 1: 2 points of intersection ∆ > 0 Case 2: 1 point of intersection Case 3: No point of intersection ∆ = 0 ∆ < 0

Determine the number of points of intersection of the parabola and the straight line. Parabola: Straight line: Example

There are two points of intersection No need to solve the eq.

Determine the number of points of intersection of the parabola and the straight line. Parabola: Straight line:

There is no point of intersection # 1

There are two points of intersection # 2

There is one point of intersection # 3

x0– 1– 2 y– 6– 30 y = – 3 x – 6 No point of intersection

x0 y0 y = 2 x – 4 Two points of intersection –

x y 2 x – y – 12 = 0 One point of intersection – – 6– 4

Determine the number of points of intersection of the parabola and the straight line. Parabola: Straight line: I) Graphical method II) Discriminant method We can use : Any other method?

Determine the number of points of intersection of the parabola and the straight line. Parabola: Straight line: I) Graphical method II) Discriminant method We can use : Any other method?

#2 ∴ (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)

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

24 – 4k = 0 # Ex.1 If the parabola and the line intersect at one point, then the discriminant equals to zero. ∴ k = 6

2.If the straight line y = 3x + k does not cut the parabola y = x 2 – 3 x + 2 at any point, find the range of values of k. Exercise There is no point of intersection

28 + 4k < 0 # Ex.2 There is no point of intersection so the discriminant is less than zero. ∴ k < – 7

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

– 3 – 12 k > 0 # Ex.3 There are two points of intersection so the discriminant is greater than zero. ∴ k < – 0.25 The largest value of k is – 1