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

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

Question 1035908: Find the equations of the lines that pass through point (-1,3) and are tangent to the curve .

Answer by Edwin McCravy(20060) (Show Source): You can put this solution on YOUR website!

Let the point of tangency be (p,q)

Then since it must satisfy y=x�-2x+1 

*  The point of tangency is (p,p�-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=x�-2x+1 at the point of tangency:
y'=3x�-2 

we substitute y'=m and x=p

m = 3p�-2

So the equation of the tangent line is now

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

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

p�-2p+1 = (3p�-2)p+(3p�-2)+3

p�-2p+1 = 3p�-2p+3p�-2+3
p�-2p+1 = 3p�-2p+3p�+1
      0 = 2p�+3p�
      0 = p�(2p+3)
      0=p�;  2p+3=0
      p=0;     2p=-3
                p=

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

y = (3p�-2)x+3p�+1
y = (3*0�-2)x+3*0�+1
y = -2x+1

We find the equation of the tangent line with p=








   

Edwin

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