Question 1035907
Let (a,b) be a point of tangency at the curve y, and let y - 3 = m(x+1) be the line of tangency passing through (-1,3), where m is the slope of the line.

==> {{{m = (b-3)/(a+1)}}} is the slope of the tangent line.

Now y' = {{{3x^2 - 2}}}, and the slope of the tangent line at the point of tangency should be m = {{{3a^2 - 2}}}.

==> {{{3a^2 - 2 =  (b-3)/(a+1)}}}

<==> {{{(3a^2 - 2)(a+1) =  b-3}}}  <==>  {{{(3a^2 - 2)(a+1) =  a^3 - 2a +1-3}}}

since (a,b) is a point on the curve.

==> {{{2a^3+3a^2 = 0}}}  after simplification.

<==>{{{a^2(2a+3) = 0}}}  ==> a = 0 or a = -3/2.

==> b = 1 or b = 5/8, respectively, after substitution.

Thus the two points of tangency are (0,1) and (-3/2,5/8).

==> There are two tangent lines.

The tangent line passing through (0,1) and (-1,3) is {{{highlight(y = -2x +1)}}}.

And, the tangent line passing through (-3/2,5/8) and (-1,3) is {{{highlight(4y = 19x +31)}}}.