Application of the second order difference quotient

Suppose the following points lie on the graph of a function $f$:
$$(0.9,0.99), (1,1), \text{ and } (1.1,0.99).$$
Use the second order difference quotient to estimate $f''(1)$.

By the second order difference quotient, $f^{''}(x) = \dfrac{f(x+h) - 2f(x) + f(x-h)}{h^{2}}$. Note $h=0.1$ in this example. Thus $$f^{''}(1) = \dfrac{f(1.1)-2f(1)+f(0.9)}{(0.1)^{2}} = \dfrac{0.99-2+0.99}{0.01} = -2$$

It may be obvious, but how do you obtain your value for h?

@sgerall $h$ is the common distance between consecutive $x$ coordinates. Thus,
$$h=1-0.9=1.1-1=0.1.$$