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

Mad Hat SG

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 SG

Hi,

Thanks for sharing these counterexamples :)

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

Cheers,
Wen Shih

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 SG

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.