Classify the discontinuity of r(x) at x = 2

Factor the numerator using the difference of cubes formula: $x^3 - 8 = (x-2)(x^2+2x+4)$. For $x \neq 2$, this lets us cancel the common factor of $(x-2)$, giving $r(x) = x^2 + 2x + 4$ whenever $x \neq 2$.

Since $r(x)$ agrees with the polynomial $x^2+2x+4$ for all $x$ near but not equal to $2$, the limit as $x\to 2$ exists and equals $2^2+2(2)+4 = 12$. Because $r(2)$ is undefined but the limit exists (and is finite), this is a removable discontinuity — the graph has a hole at $(2,12)$ that could be patched by defining $r(2)=12$.

The graph shows a hole. Since the discontinuity at x equals two is removable, the limit exists and equals twelve, but the function value there is undefined. This creates a single missing point on an otherwise continuous curve, which is visually represented as a hole rather than a jump or a break from an asymptote.