Finding equation of tangent

• fireice rox

  18 Nov 09, 18:22

  Hi, need help in solving this:

  The tangent to the curve y = x^2 - 2x + 3 at a certain point is parallel to the line y = x. Find the equation of the tangent

• FireIce

  18 Nov 09, 18:28

  dy/dx = 2x -2
  
  gradient of y = x is 1
  // lines have same gradient

  sub in.

• Okmijnub

  18 Nov 09, 18:42

  dy/dx= 2x - 2

  when dy/dx = 1 , x = 1.

  Sub x = 1 into y = x^2 - 2x + 3, so y = 2.

  Sub Y=2 and  x=1 into y-y(1)=1(x-x(1)) ,

  You will get-----> y = x + 1

  PS: Just now I do wrongly.

• TrueHeart

  18 Nov 09, 22:53

  Originally posted by Okmijnub:

  dy/dx= 2x - 2

  when dy/dx = 1 , x = 1.

  Sub x = 1 into y = x^2 - 2x + 3, so y = 2.

  Sub Y=2 and  x=1 into y-y(1)=1(x-x(1)) ,

  You will get-----> y = x + 1

  PS: Just now I do wrongly.

  Hi Okmijnub, 

  There is a slight mistake in your workings.

  When dy/dx = 1, x = 3/2. Hence, y = 9/4.

  The equation of the tangent is :

  1 = [y - 9/4] / [x - 1]

  y - 9/4 = x - 1

  y = x + 5/4

  Nevertheless, it is very kind of you to help out other forummers in need. It is the thought that matters the most.

  Cheers.

• deepak.c

  18 Nov 09, 22:54

  Originally posted by TrueHeart:

  Hi Okmijnub, 

  There is a slight mistake in your workings.

  When dy/dx = 1, x = 3/2. Hence, y = 9/4.

  The equation of the tangent is :

  1 = [y - 9/4] / [x - 1]

  y - 9/4 = x - 1

  y = x + 5/4

  Nevertheless, it is very kind of you to help out other forummers in need. It is the thought that matters the most.

  Cheers.

   

  I got 9/4 too.