differentiation

a) As the curve passes through the origin (0,0),
9*0=a(0)^3+b(0)^2+c(0)+d
d=0
y= (a/9)x^3+(b/9)x^2+(c/9)x
dy/dx=3(a/9)x^2+2(b/9)x+(c/9)
=(a/3)x^2+(2b/9)x+(c/9)
As the tangent at (0,0) makes an angle of 45' with the +ve x-axis, dy/dx|x=0 =1.
dy/dx|x=0 =(a/3)(0^2)+(2b/9)x+(c/9)
1=c/9
c=9
As the x-axis is a tangent to the curve at x=3, y=0 when x=3 and dy/dx|x=3 = 0
0=(a/9)(3^3)+(b^9)(3^2)+(9/9)(3)
3a+b+3=0
dy/dx|x=3 = (a/3)(3^2)+(2b/9)(3)+(9/9)
0=3a+(2b/3)+1
9a+2b+3=0
(9a+2b+3)-2(3a+b+3)=0-2(0)
3a-3=0
a=1
3(1)+b+3=0
b=6
Therefore, a=1, b=6, c=9 and d=0.
b) dy/dx=(1/3)x^2+(2*6/9)x+(9/9)
=(1/3)x^2+(4/3)x+1
Set dy/dx = 0
(1/3)x^2+(4/3)x+1=0
x^2+4x+3=0
(x+1)(x+3)=0
x+1=0 or x+3=0
x=-1 or x=-3
d^2y/dx^2=(2/3)x+(4/3)
d^2y/dx^2|x=-1 = (2/3)(-1)+(4/3)
= 2/3 >0
d^2y/dx^2|x=-3 = (2/3)(-3)+(4/3)
= -2/3 <0
y=(1/9)x^3+(2/3)x^2+x
When x=-1, y=(1/9)(-1)^3+(2/3)(-1)^2+(-1)
=-1/9+2/3-1
=-4/9
When x=-3, y=(1/9)(-3)^3+(2/3)(-3)^2+(-3)
=-3+6-3
=0
Therefore, the turning points of C are (-1,-4/9) and (-3,0).
Set d^2y/dx^2 = 0
(2/3)x+(4/3)=0
x=-2
When x=-2, y=(1/9)(-2)^3+(2/3)(-2)^2+(-2)
=-8/9+8/3-2
=-2/9
As d^2y/dx^2 < 0 when x<-2 and d^2y/dx^2 > 0 when x>-2, the point of inflexion of C is (-2,-2/9).

As for the curve sketching of part c), I am afraid you will have to refer to a graph sketching software. Search Graphmatica on Google. It is recommended by my teacher. To save your time browsing for the equation again, here it is: y=(1/9)x^3+(2/3)x^2+x