1) the tangent line at x=2 is y = x - 7 slope of the tangent is 1 then slope of the normal is -1 the normal is y = -x + C at x=2, y=2-7=-5 (the normal also pass this point) so -5 = -2 + C C = -3 the normal is y = -x -3 .... ans
2) slope of the tangent (=1) is the slope of the curve at x=2 given. y' = ax -5 then 1 = a(2) - 5 a = 3 .... ans y' = 3x - 5 y = int(3x - 5) = 3x^2/2 -5x + c we know at x=x, y=-5 so -5 = 3(2^2)/2 -5(2) + c c = -1 y = 3x^2/2 -5x -1 (curve eq) .... ans
