Quadratic Functions 2.

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Presentation transcript:

Quadratic Functions 2

Intersection Of Lines And Parabolas (Tangents To Curves) To find the point(s) of intersection between a parabola and a straight line , put the two equations together and rearrange to form a quadratic equation with the right -hand side = 0 . So, point(s) of Intersection given by Solving this equation will produce any point(s) of intersection.

We can also make use of previous results using the DISCRIMINANT CONDITION RESULT There are two real and distinct roots :- (a) So, two points of intersection. There is only one real root, so only one point of intersection.. (b) So , line is a tangent. There are no real roots. So there are no points of intersection. (c)

Diagrams for each of these situations would be as follows :- (a) two points of intersection. (b) one point of intersection.(Tangent) (c) No point of intersection

Example 1 SOLUTION Point of intersection when = (-6)2-(4 x 1 x 9) = 36 - 36 = 0 Since b2- 4ac = 0 there is only one point of intersection , so the line is a tangent .

To find the point of contact solve the quadratic equation that represents the intersection between the line and the parabola. In this example this is (x - 3)(x - 3) = 0 x = 3 Substitute this x-value into either the parabola equation or the line equation to find the y-coord. of the point of contact. y = 15 - 7 x 3 = 15 - 21 = - 6 The point of contact is ( 3 , -6)

Example 2 The line y = -2x + k is a tangent to the parabola y = 4x - x2 . Find the value of k. Solution y = -2x + k meets y = 4x - x2 where -2x + k = 4x - x2 x2 - 6x + k = 0 Since the line is a tangent b2- 4ac = 0 (-6)2 - (4 x 1 x k) = 0 36 - 4k = 0 4k = 36 k = 9 So, the equation of the tangent is y = -2x + 9 OR y = 9 - 2x