Finding points at which line tangent to curve $y = x^3 - 3x^2 - 2x + 1$ forms equal angles to $x$- and $y$-axes

Question

$y = x^{3}-3x^{2}-2x+1$

It's a curve line. I have to find out the points, by which the tangent go through in such a way that, these tangents create equal angles to the both $x$- and $y$-axes. More precisely ,

Tangent will go through some points, so that, tangent create same equal angle to the both axes. I need to find out the points.

Thanks.

Answer 1

If you want to have a view, draw coordinates, the line passing point A(a, 0) and B(0, a) has equal angles(45^o) with x and y axis.For formulation the gradient of line is:

$m= tan(180- 45=135)=-1$

and equation of line is $y=-1(x-1)$ or $y=-x+1$.

For your question m=-1 is the gradient of tangents to curve on first and third quarter of coordinate system. For second and forth quarter m=1.Now take derivative of function and equate it to -1, you find a quadratic equation, solving it give the x of required points. Plug them in equation you get y of the points.These are points for tangents in first and third quarters.Do the same with m=1 to find points of tangents on second and forth quarter.