G. Tawfik          
MATHEMATICS 436

Chap. 4C Review: Linear and Quadratic Problems Functions

Notes: 1. Linear function
                2.  Quadratic function

     1.    The middle cable of a suspended bridge has the shape of a parabola. The distance between the two supporting towers is 120 m.
            The distance between Vertex V and point M midpoint between the tower tops is 25 m.
            How high above the bridge span is a flag F that is 20 m away from vertex V?                              



    2.    An archway has the shape of a parabola as shown. The base of the arch is 60 m wide.
           Points A and B 15 m high are 44 m apart.
           How high is this arch?

  




  1. In right triangle ABC, the equation of side AB is: x - 4y + 16 = 0, vertices B and C are on the y-axis                                                                and the coordinates of vertex A are ( -4, 3). Determine the numerical value of the area of triangle ABC.





  1. Pam (P) and Elyse (E) are playing with a ball in the swimming pool.  The following diagram shows the parabolic trajectory                           of the ball thrown by Pam.

 

 

        What is the rule of correspondence of this trajectory if point P is at the origin of the coordinate system, point B is the maximum height
        of the ball and point E has coordinates (6, 0)?

 

     5.    An engineer sketched a parabola in the Cartesian plane.

 

 

        Which rule of correspondence defines this parabola?

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6.      A study compared the number of members in two health clubs over a 30‑day period. The following graph represents this situation.

 

 

        The change in the number of members in Club A over this period is represented by a portion of a parabola.
        The coordinates of the vertex of this parabola are (0, 150).

        On the 30th day, what was the exact difference between the number of members in Club A and the number of members in Club B?

 



 



R4. Review C  Answers:


 1)      (0, 0)   (120, 0)  V (60, -25)   y = 0,007(x – 60)2 – 25   F (40, -22,2) = 2.8 m high

 

2)      (0, 0)   (60, 0)  A (8, 15)         y = - 0,036x(x – 60)   V (30, 32,4)

 

3)      A ( -4, 3)   B (0, 4)    C (0, 13)     Area = 34 m2 

 

4)      P (0, 0)     B (3, 4)     C ( 6, 0)     y = - 4/9  (x – 3)2 + 4

 

5)      A (2, 0)   B (14, 0)    V (8, -6)     y =  1/6  (x – 8)2 - 6