H2 maths: vertical tangent issue for graph of y^2 = f(x)

• Mad Hat

  26 Mar 13, 00:10

  According to C.S. Toh (A level study guide), for the graph of y² = f(x):

  Where the graph crosses the x-axis, the gradient of the graph is infinity, i.e., the tangent is vertical. (p. 17)

  Many students appeared not to know the features of the graph of y² = f(x), e.g. that its gradient when it crosses the x-axis is infinite, i.e. the tangent is vertical. (p. 26)

  Some of my tuition students tell me that their teachers say the same thing - the tangent at the x-axis should always be vertical for the graph of y² = f(x).

  Where does this vertical tangent thing come from? It seems to be a piece of misinformation. It is true that, for many graphs, the vertical tangent is present. But there are equally many graphs where there is no vertical tangent at the x-axis at all.
  
  E.g. suppose the original graph is y = x² (red graph) Now we sketch y² = x² (blue graph):

  

  Obviously, there is no vertical tangent at the x-axis. 

  Same thing with y = x³ (red graph) and y² = x³ (blue graph). There is no vertical tangent at the x-axis:

  

  Any idea why this "vertical tangent" misinformation is going around?

• wee_ws

  26 Mar 13, 08:35

  Hi,

    Thanks for sharing these counterexamples :)

    Students simply rote-learn and teachers/books simplify matters too much :P

  Cheers,
  Wen Shih

• wee_ws

  26 Mar 13, 08:51

  Hi,

    Given: y^2 = f(x)

    => 2y y' = f'(x)

    => y' = f'(x) / {2 sqrt(f(x))}.

    We may remind students these points:

    (i) If f(x) is a perfect square, then we sketch the graph of y = +/- |sqrt(f(x))|.

    Example: y = x^2, we sketch y = +/- |x| when we draw the graph of y^2 = x^2.

    (ii)  If y = f(x) has a stationary point at x = a, then y' = 0.

    Example: y = x^3 has a stationary point at x = 0, so y' = 0 at x = 0 when we sketch the graph of y^2 = f(x).

    (iii) If y = f(x) has no stationary point at x = a, then y' is undefined.

    Example: y = x^3 + x has no stationary point at x = 0, so y' is undefined at x = 0 when we sketch the graph of y^2 = f(x).

    What do you think? Thanks in advance!

    Thanks.

  Cheers,
  Wen Shih
  wenshih.wordpress.com

• Mad Hat

  26 Mar 13, 18:09

  Wen Shih, thanks, that equation y' = f'(x) / 2√f(x) is useful !

  So, from what I can see, the tangent for y² = f(x) may be assumed to be vertical at the x-axis provided the original graph does not have a stationary point there.

  This is something I can tell my students. Easy to digest.

  I'm not comfortable with point (ii) however, cos, if both f(x) and f'(x) = 0, then y' = 0/0, which is undefined also? I think I prefer to say that if the original graph is flat at the x-axis, then the tangent for y² = f(x) at that point could be anything, i.e., no obligation to draw it as vertical, especially if it's not natural when drawing the graph.