The equation of a curve is y = ax^2 + 3x + c where a and b are integers. The curve has a minimum point at (1,1), find a and c
We start by trying to find out the values of a and c using the information about the minimum point. We know we can rearrange the right hand side by completing the square: y = a(x^2 + 3x)+cy = a(x+3/2a)^2+c-(9/(4a)) as y = a(x^2 + (3/a)x + 9/(4a2)) = ax^2 + 3x + 9/4a to ensure that this is equivalent to ax^2 +3x + c, we substract 9/4a and add c. we know that this is a minimum point so we need the square to be minimum as if the inside is negative, negative squared is positive and if it is positive, positive squared is positive the minimum must be where the square is 0. So x = -3/(2a) so 1 = -3/2a so a = -3/2 as -3/(2*(-3/2)) = 1. We now know that 1 = 0+c-(9/4(-3/2)), 1 = c + (9/6) so c = -3/6 .
