Saying [math]f[/math] is differentiable at [math]x=a[/math] means that [math]f'(a)[/math] (which is defined as [math]\lim_{x\to a}\frac{f\left(x\right)-f\left(a\right)}{x-a}[/math] or, equivalently, [math]\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)}{h}[/math]) exists (and is a real number).

The point-slope form of a line is [math]y-y_1=m\left(x-x_1\right)[/math]. This line passes through [math]\left(x_1,y_1\right)[/math] and has slope [math]m[/math]. The line [math]y=f'\left(a\right)\cdot\left(x-a\right)+f\left(a\right)[/math], which can also be written [math]y-f\left(a\right)=f'\left(a\right)\cdot\left(x-a\right)[/math], is therefore a line passing through [math]\left(a,f\left(a\right)\right)[/math] (which is the intended point of tangency) with slope [math]f'\left(a\right)[/math] (which is the slope of the tangent line).

Over a small neighborhood of [math]a[/math], no other line passing through [math](a,f(a))[/math] deviates from [math]f[/math] less than the tangent line. That is, tilting this tangent line in either direction would result in more deviation between the line and [math]f[/math].

At [math]x=0[/math], GeoGebra fails to plot a tangent line. The function [math]f\left(x\right)=x^{\frac{1}{3}}[/math] is not differentiable at [math]x=0[/math]. Thus, according to Good Definition #1, no tangent line exists. However, there is a best linear approximation of [math]f[/math] at [math]x=0[/math] (namely, the vertical line [math]x=0[/math]), so Good Definition #2 would say a tangent line does exist.