190 CHAPTER 3 Solution. Then F (x) = (6, -12). Therefore, the solution space of F (x)x = 0 m is the set of vectors (2t, t) T , t R, and its orthog- onal complement is the set of vectors (s, -2s) T , s R. To find the quadratic approximation, we have to solve the following equation in the variable c: F (3 + 2t + ct 2 , 2 + t - 2ct 2 ) = o(t 2 ) for all t R. This gives, (3 + 2t + ct 2 ) 2 - (2 + t - 2ct 2 ) 3 - 1 = (4t 2 + 6ct 2 ) - (6t 2 - 8ct 2 ) = (14c - 2)t 2 = o(t 2 ), and hence c = 1/7. Therefore, the quadratic approximation of x 2 - x 3 - 1 = 0 1 2 at the point x = (3, 2) T has the following parametric description 2 1 2 2 T