Question 1153034
x^2y-x=y^3-8 at x=0
<pre>
Substituting x=0,
{{{x^2y-x=y^3-8}}}
{{{0^2y-0=y^3-8}}}
{{{0=y^3-8}}}
{{{8=y^3}}}
{{{2=y}}}

So the point of tangency is (0,2), and we want the equation of the
green line below:

{{{drawing(400,400,-4,4,-4,4,
graph(400,400,-4,4,-4,4,12,(-1/12)x+2),
locate(0,2,"(0,2)"),
graph(400,400,-4,4,-4,4,((2/3)^(1/3)x^2)/(sqrt(3)sqrt(-4x^6 + 27x^2 - 432x + 1728) - 9x + 72)^(1/3) + (sqrt(3)sqrt(-4x^6 + 27x^2 - 432x + 1728) - 9x + 72)^(1/3)/(2^(1/3)3^(2/3) )))}}}

We find the slope of the tangent line, which is the
same as the derivative at that point:  So we find
the derivative implicitly, i.e., without solving for
the independent variable y:

{{{x^2y-x=y^3-8}}}
{{{x^2*expr(dx/dy)+2xy-1=3*y^2*expr(dy/dx)}}}

We substitute x=0 and y=2 and solve for {{{"dy"/"dx"}}}

{{{0^2*expr(dx/dy)-2(0)(2)-1=3*2^2*expr(dy/dx)}}}

{{{0-0-1=3*4*expr(dy/dx)}}}

{{{-1=12*expr(dy/dx)}}}

{{{-1/12=expr(dy/dx)}}}

That's the slope of the tangent line at (0,2), which is
the green line. So

{{{m=-1/12}}}

Since the point of tangency is the y-intercept (0,2), we can just
use:

{{{y=mx+b}}}

with 
{{{matrix(1,3,m=-1/12,"", b=2)}}}

{{{y=expr(-1/12)x+2}}}

Edwin</pre>