Find the point on the parabola y = :2 that is the closest to the point (6.. 6} on the Cartesian plane. at Write the formula for the square of the distance between a point (1,:2} on the parabola and the point (6.6). Answer: h{:] = [xB}"2+()c\"2-6}\"2 @@ b} Compute the derivative of h. Answer: hire) = 4*x-'322'x-12 @l c: Find the critical points p1 -=:: p2. z: p3 ofh. Give your answers with an accuracy of two decimal places. Answer: p1 = 42.00 p2 = 43.58 p3 = 2.58 :1} Suppose that Pl <:: p2 a: p3 are the three critical points that you have found in complete following table. for fields which to provide a numerical value give your answers with an accuracy of two decimal places. range ol : behaviour m2 p1 strictly decreasing z="p1" local minimum v p c: increasing i :: e: what is point ofthe parabola closest and this distance don forget ms represents square distance. answer: y d must use braclcets instead parentheses denote vectorsi. n.ir>