Answer: B
Step-by-step explanation:
[tex]f(x) = \dfrac{x^2-16}{x-4}[/tex]
Restriction: denominator cannot equal zero so x - 4 ≠ 0 → x ≠ 4
[tex]f(x) = \dfrac{(x-4)(x+4)}{x-4} = x + 4[/tex]
Since the denominator (x - 4) cancelled out, x = 4 is not a vertical asymptote - it is a hole.
f(x) = x + 4 is the simplified function
f(4) = (4) + 4
= 8
So, the hole is at (4, 8)
The graph is the line y = x + 4 with a hole at (4, 8). The only graph that has these attributes is the top right graph, which I call graph B.
