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

Algebra ->  Points-lines-and-rays -> SOLUTION: Find the equations of the lines that pass through point (-1 , 3) and are tangent to the curve {{{ y=x^3 - 2x + 1}}} . Graph the tangents and the curve      Log On


   

Question 1035907: Find the equations of the lines that pass through point (-1 , 3) and are tangent to the curve +y=x%5E3+-+2x+%2B+1 .
Graph the tangents and the curve
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
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+=+%28b-3%29%2F%28a%2B1%29 is the slope of the tangent line.
Now y' = 3x%5E2+-+2, and the slope of the tangent line at the point of tangency should be m = 3a%5E2+-+2.
==> 3a%5E2+-+2+=++%28b-3%29%2F%28a%2B1%29
<==> %283a%5E2+-+2%29%28a%2B1%29+=++b-3 <==> %283a%5E2+-+2%29%28a%2B1%29+=++a%5E3+-+2a+%2B1-3
since (a,b) is a point on the curve.
==> 2a%5E3%2B3a%5E2+=+0 after simplification.
<==>a%5E2%282a%2B3%29+=+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%28y+=+-2x+%2B1%29.
And, the tangent line passing through (-3/2,5/8) and (-1,3) is highlight%284y+=+19x+%2B31%29.