Question 350591
{{{y=Ax^3+Bx^2+Cx+D}}}
The slope of the tangent line at a given point is the value of the derivative at that point. 
The derivative of the function is,
{{{dy/dx=3Ax^2+2Bx+C}}}
At x=0, the slope of the tangent line is {{{m=1}}}.
{{{m=dy/dx(0)=3A(0)+2B(0)+C=1}}}
{{{highlight(C=1)}}}
.
.
.
At x=2, the slope of the tangent line is {{{m=2}}}.
{{{m=dy/dx(2)=3A(2)^2+2B(2)+1=2}}}
{{{12A+4B+1=2}}}
1.{{{12A+4B=1}}}
.
.
.
Using the tangent line at x=0, you also know the intersection point because,
{{{y=x}}}
{{{y=0}}}
(0,0) is also a point on the cubic.
{{{y=Ax^3+Bx^2+Cx+D}}}
{{{y=A(0)+B(0)+(0)+D=0}}}
{{{highlight(D=0)}}}
.
.
.
Similarly at x=2,
{{{y=2x-3}}}
{{{y=2(2)-3}}}
{{{y=1}}}
.
.
{{{y=Ax^3+Bx^2+x}}}
{{{1=A(2)^3+B(2)^2+2}}}
{{{1=8A+4B+2}}}
2.{{{8A+4B=-1}}}
Subtract eq. 2 from eq. 1,
{{{12A+4B-8A-4B=1-(-1)}}}
{{{4A=2}}}
{{{highlight(A=1/2)}}}
Then from eq. 2,
{{{8(1/2)+4B=-1}}}
{{{4+4B=-1}}}
{{{4B=-5}}}
{{{highlight(B=-5/4)}}}
.
.
.
{{{drawing(300,300,-3,5,-10,10,circle(0,0,0.15),circle(2,1,0.15),graph(300,300,-3,5,-10,10,(1/2)x^3-(5/4)x^2+x,x,2x-3))}}}