Which of the following best explains why the graph of contains a hole at rather than…
MATHEMATICS · Single correct · AP · AP Precalculus · Polynomial and Rational Functions · Polynomial and Rational Functions
Problem
Which of the following best explains why the graph of f(x)=x−3x2−9 contains a hole at x=3 rather than a vertical asymptote?
(a) Because the denominator approaches zero more quickly than the numerator as x→3, the quotient becomes unbounded, producing a vertical asymptote.
(b) Because the factor (x−3) cancels completely, the simplified function equals x+3 for all x, so the graph is actually continuous at x=3.
(c) Because (x−3) is a common factor in both numerator and denominator, it cancels, leaving a finite limit as x→3 even though the function is undefined there; this creates a removable point (hole) rather than a vertical asymptote.
(d) Since both numerator and denominator are zero at x=3, the ratio equals 0/0=0, so the graph merely has a missing point on the x–axis at that value.