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
