For the smooth curve x(t) = 4t2, y(t) = t3 4t, nd all points where the tangent line is either horizontal or vertical. (Use symbolic notation and fractions where needed. Give your answers in the form of comma separated lists of point coordinates in the form (*, *). Enter DNE if there are no such points.) I have checked many times, the solution given by me , as well as by graphically. Parametric equations acct) = 4+2 and yet)= + 3-4+ y = t (t 2-4) Then 42 = +2 ( t ?-4) 2 substitute DC = 462 - 1= 4 2 = i ( 2 - 4 ) 2 y? = * ( x- 16 ) = = 25. (x-16)2 y ' = 36 (20 - 16 ) 2 64 64 42 = DC (DC-16)2 find dy doc 64( 24) . dy = 1. (20-16) + 2C. 2(20-16) dy doc = ( 30-16) ( 20-18 + 206) y's = = ( xx - 16) 2 dy /20 - 16 doc 64 . 24 ) (320- 18) 8 Y = JOE (DC-16) dy 8 dac = (320-16) 128 , 525 (20- 5) IC dy 320- 16 when =0, 32-16 =0 doc 16JJC DC = 16 = 5.3333 when dy = = 320- 16 - = 5X= 0, 2C = 0 Sowe find ac = 18 and corresponding value of y = + 1653= 4 3. 079 for horizontal tangent 9 and for vertical tangent sc = 0 andy = 0 I include two graph of function y? = 24 (x-16)2 and tangent lines