Question 2032
 If a given quadratic equation as:

 y = ax^2 +b x + c.  (Note a <> 0,& b,c reals)

 To find x-intercept,since the corresponding point which y-coordinate
 must be 0. We have to solve the equation

 y = ax^2 +b x + c = 0...(*).
 
 In general,if the discriminant b^2 -  4ac > 0, (*) has two different
 real roots and hence there are two x-intercepts 
 (-b+sqrt(b^2 -  4ac))/2a or  (-b-sqrt(b^2 -  4ac))/2a if b^2 -  4ac > 0.

 [Note: IF b^2 -  4ac = 0 , there is only one x-intercept -b/2a
        IF b^2 -  4ac < 0 , there are no x-intercepts ]

 To find y-intercept,since the corresponding point which x-coordinate
 must be 0. When we set x =0, we obtain
 y = c. Hence, there is only one y-intercept c.


 Kenny