SOLUTION: Find the equations of the lines that pass through point (-1,3) and are tangent to the curve {{{ y=x^3 - 2x + 1 }}} .

Let the point of tangency be (p,q)

Then since it must satisfy y=x³-2x+1 

*  The point of tangency is (p,p³-2p+1)

Let the equation of the tangent line be y = mx+b

Since the tangent line goes through (-1,3),

    3 = m(-1)+b
    3 = -m+b
  3+m = b

*  Equation of tangent line is y = mx+m+3

The slope of the tangent line is the derivative of
y=x³-2x+1 at the point of tangency:
y'=3x²-2 

we substitute y'=m and x=p

m = 3p²-2

So the equation of the tangent line is now

    y = mx+m+3
    y = (3p²-2)x+(3p²-2)+3
*   y = (3p²-2)x+3p²+1

Since the tangent line must go through the point
of tangency (p,p³-2p+1)

p³-2p+1 = (3p²-2)p+(3p²-2)+3

p³-2p+1 = 3p³-2p+3p²-2+3
p³-2p+1 = 3p³-2p+3p²+1
      0 = 2p³+3p²
      0 = p²(2p+3)
      0=p²;  2p+3=0
      p=0;     2p=-3
                p=

We find the equation of the tangent line with p = 0

y = (3p²-2)x+3p²+1
y = (3*0²-2)x+3*0²+1
y = -2x+1

We find the equation of the tangent line with p=








   

Edwin