The Mean Value Theorem asserts that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least one $c\in (a,b)$ such that ${\displaystyle f'(c) = \frac{f(b)-f(a)}{b-a}}$. We can restate the conclusion in graphical terms as follows. There must be a tangent line to the graph of $f$ which is parallel to the secant through $(a,f(a))$ and $(b,f(b))$ with the point of tangency between $(a,(f(a))$ and $(b,f(b))$. Equivalently, if at each $c\in (a,b)$ we construct the line through $(c,f(c))$ parallel to the secant through $(a,f(a))$ and $(b,f(b))$, then at least one of these lines must be tangent to the graph of $f$ at $(c,f(c))$.
The Sage interact on this page allows you to enter an expression for $f(x)$ and numerical values for $a$ and $b$. It plots $f(x)$ on the interval $[a,b]$ (actually a slightly larger interval) and the secant through $(a,f(a))$ and $(b,f(b))$. The interact also plots a line parallel to this secant through a point $(c,f(c))$ where you may control $c$ using the slider $r$. (The value of $r$ specifies the fraction of the distance from $a$ to $b$ to travel from $a$ [to the right] to get to $c$.) If $f$, $a$ and $b$ satisfy the hypotheses of the mean value theorem, then there should be at least one slider position which generates a tangent to the graph of $f$. If you perceive the inevitability of success (within the limitations of our digitized model), then you have an intuitive grasp of the mean value theorem.