Non-differentiable points in a graph: Vertical Tangent vs Vertical Asymptote Can someone help me understand the difference between the two?
Vertical tangent at $x = a$: $f$ is continuous at a but $f'(a)$ blows up.
· How is this different from a vertical asymptote?
 A: For asymptote f is not continuous. think tor exaple tangent function
A: It is possible for a graph to have a vertical tangent.  For example the function $y=\sqrt[3]x$ has the vertical tangent $x=0$ even though its slope $y=dy/dx$ is undefined. Also, the unit circle, defined parametrically as $(\cos t, \sin t)$ has a vertical tangent at $t=n\pi$ for all $n\in\Bbb Z$ even though its slope $(dy/dt)/(dx/dt)$ is undefined.
A vertical asymptote, on the other hand, is a line of the form $x=a$ where $a\in\Bbb R$ that a function tends to. The function could simply approach the asymptote from either side but not be defined there, as with $y=1/x$, or it could swivel around the asymptote, as with the parametric function $\left( \cos(t)/t , t \right)$.
A: If you carefully compare the definitions, you will see that $f$ has vertical tangent at $a$ when $f'$ has vertical asymptote at $a$. Note that $f$ does not have vertical asymptote at $a$ in that case, since by $f$ "blowing up", we would immediately have that $f$ is discontinuous at $a$.
A: Two examples:
$$ y= \frac{1}{a-x}$$
has a vertical asymptote at $x=a$, the curvature of the curve goes to zero but slope is infinite. In Control engineering called a pole.
$$ y^2= {2 a x-x^2}$$
has a vertical tangents at points $ (x=0, x=2 a) $ on x-axis. Continuously differentiable.
The curvature of the curve remains at $ \dfrac{1}{a},$ and these points are zeroes.
