Since the numerator is not zero when $ t = 2 $, and the denominator is zero, there is a vertical asymptote at $ t = 2 $. There are no common factors between numerator and denominator, so there are no holes.
["# Understanding Vertical Asymptotes in Rational Functions: Why $ t = 2 $ Creates One in Specific Cases", "When analyzing rational functions—expressions of the form $ \frac{P(t)}{Q(t)} $, where $ P(t) $ and $ Q(t) $ are polynomials—vertical asymptotes occur at values of $ t $ that make the denominator $ Q(t) = 0 $, provided the numerator $ P(t) $ does not equal zero at those points. One classic example occurs when $ t = 2 $ causes a vertical asymptote despite the numerator being non-zero.", "## What Causes a Vertical Asymptote at $ t = 2 $?", "By definition, a vertical asymptote arises at $ t = a $ if:", "- The denominator $ Q(a) = 0 $, meaning the function is undefined at that point.\n- The numerator $ P(a) <br/>\ne 0 $, ensuring the function approaches infinity or negative infinity rather than a removable discontinuity (a "hole").", "The presence of a vertical asymptote at $ t = 2 $ in a rational function means $ Q(2) = 0 $, but when we substitute $ t = 2 $ into $ P(t) $, the result is nonzero. Since no common factors exist between $ P(t) $ and $ Q(t) $, the discontinuity at $ t = 2 $ cannot be canceled—it is a true asymptotic behavior.", "### Why Vertical Asymptotes—not Hole—Occur at $ t = 2 $", "When the numerator is nonzero at $ t = 2 $, the function grows without bound as $ t $ approaches 2 from either side. This behavior matches the formal definition of a vertical asymptote. Crucially, since $ Q(2) = 0 $ and $ P(2) <br/>\ne 0 $, the function has a singularity at $ t = 2 $, but nowhere else in the domain. No hole forms because the zero in the denominator is simple—not canceled by a matching zero in the numerator.", "## How to Identify This Situation in Practice", "Consider a rational function such as:", "$$\nf(t) = \frac{t - 3}{(t - 2)^2}\n$$", "At $ t = 2 $, the denominator becomes zero: $ (2 - 2)^2 = 0 $, and the numerator is $ 2 - 3 = -1 <br/>\ne 0 $. Since the numerator is nonzero and no factor cancels $ (t - 2) $ in the denominator, $ t = 2 $ is a vertical asymptote, not a hole.", "By contrast, if both numerator and denominator were zero (e.g., $ \frac{t - 2}{(t - 2)(t + 1)} $), cancellation may produce a removable discontinuity (a hole) at $ t = 2 $, but no asymptote.", "## Key Takeaways for Students and Enthusiasts", "- A vertical asymptote exists where denominator is zero and numerator is nonzero.\n- No common factors between numerator and denominator prevent cancellation and guarantee a true asymptote.\n- Avoid confusing vertical asymptotes with removable discontinuities or holes by checking zeros and factorization.\n- Analyzing vertical asymptotes enhances understanding of function behavior, limiting, and graphing.", "Recognizing vertical asymptotes helps not only in pure math but also in engineering and science applications where system stability and response to inputs depend on function behavior near critical points.", "---", "Identifying vertical asymptotes is essential for mastering rational functions. When the numerator is safe—nonzero at $ t = 2 $—the denominator’s zero marks a point where the graph shoots off to infinity, clearly outlining the function’s limits at boundaries within its domain."]
