The absolute function $$f(x)=\begin{cases} x, &\text{if $x\ge0$}\\ -x, &\text{if $x<0$} \end{cases}$$
has the derivative
$$f'(x)=\begin{cases}
1, &\text{if $x\ge0$}\\
-1, &\text{if $x<0$}
\end{cases}$$
So the slope of the graph at $x=0$ is
$$f'(0)=1$$
The graph of $f(x)$ and it's slope (calculated by $f'(x)$) at $x=0$ are shown below.
However I was wondering that the slope of $\vert x\vert$ at $x=0$ should be $0$, based on the fact that the tangent should be the $x$-axis as shown below.
My question is: Shouldn't the value of the slope of $\vert x\vert$ at $x=0$ be $0$ based on graphical methods. I say this because slope is rather a graphical concept while differential is the calculus concept. If the slope is $0$, what method should be used to derive this result?
First, in $x = 0$, we have:
$$\lim_{x\to0-}\frac{|x|-0}{x-0} = \lim_{x\to0-}\frac{-x}{x} = -1$$
On the other hand,
$$\lim_{x\to0+}\frac{|x|-0}{x-0} = \lim_{x\to0+}\frac{x}{x} = 1$$
Because both limits are not the same, it follows that the function defined by $f(x) = |x|$ is not differentiable at $x = 0$ (this is the calculus approach to see that the slope is not defined)
If you know the geometrical meaning of the derivative, you should be able to see why the slope of the tangent line in a point at the graph and the derivative are the same thing. Intuitively, you can think about this in the following way:
Remember, $$f'(a) = \lim_{x\to a}\frac{f(x) - f(x)}{x-a}$$
so you can think about the quotient $\frac{f(x) - f(x)}{x-a}$ where $x$ is a number really close to $a$. You will notice, when you make a drawing, that this is more or less the slope of the tangent line.
Now, to answer your question, it should be clear that the slope is not defined at $x = 0$. How would you draw a unique tangent line? As mentioned in the comments, symmetry is no reason. If you consider $y = 0$ to be the tangent line, then other lines such as $y = x$ and $y = -x$ are also tangent lines.