Suppose that a parabola y=a x^2+b x+c , where a 0 and (a+b+c) is an integer has vertex ( 1 4 , -9 8 ) . If the minimum possible value of 'a' can be written as p q where p and q are relatively prime positive integers, then find (p+q)

Step-by-step solution

The equation of the parabola can be taken as y=m (x- 1 4 )^2- 9 8 (given that minimum value is -9 8 when x= 1 4 ) or y=m (x^2+ 1 16 - x 2 )- 9 8 or y = mx ^2- m 2 x + m 16 - 9 8 .....(1) Comparing equation (1) with y=a x^2+b x+c , Hence a=m, b= -m 2 and c= m 16 - 9 8 But a+b+c is an integer m 2 + m 16 - 9 8 = 9 m-18 16 is an integer. If a+b+c=0 m=2 If a+b+c=-1 m= 2 9 If a+b+c=-2 m= -14 9 which is negative Hence minimum positive value of coefficient of x^2 which is m= 2 9 p=2 and q=9 Hence (p+q)=11