The top gives the zeroes. The bottom gives the walls.
Find the vertical asymptote
f(x) = x − 3x + 2
A vertical asymptote sits where the bottom is zero and the top is not. Here x + 2 = 0 at x = −2, and the top at that point is −5, not 0. A is the zero, where the graph crosses the x-axis, not a wall. C is the horizontal asymptote — a real feature of this graph, but the question asked for the vertical one. The sign trips people up: x + 2 is zero at negative 2.
y-intercept: f(0) = −32
put x = 0
x-intercept: x − 3 = 0
only the TOP can make it zero
x = 3
the zero of the function
A fraction is zero only when its numerator is zero. Nothing you do to the bottom can produce a zero.
Close to an asymptote the outputs run away to ±∞. The curve never touches the line, which is what makes it an asymptote rather than an intercept.
For x − 3x + 2 both degrees are 1 and both leading coefficients are 1, so the horizontal asymptote is y = 1.
x² − 4x − 2, x ≠ 2
restriction first
= (x + 2)x − 2x − 2
factor
= x + 2, but only for x ≠ 2
a line with one point missing
The graph is the straight line y = x + 2 with a hole punched at (2, 4). Writing the answer as "y = x + 2" and stopping loses the one fact that made the question interesting.
Each feature tells us one piece: a zero is a factor on top, a wall is a factor on the bottom, and the horizontal asymptote fixes the leading coefficients.
Zero at x = −2, vertical asymptote x = 5, horizontal asymptote y = 3.
top has (x + 2)
zero at −2
bottom has (x − 5)
wall at 5
f(x) = 3(x + 2)x − 5 = 3x + 6x − 5
the 3 makes the ratio 31
f(−2) = 0 ✓, bottom 0 at 5 ✓, 31 = 3 ✓
check all three
The signs flip on the way in: a zero at −2 is the factor x + 2, a wall at 5 is x − 5.
Does this graph ever reach y = 3?
3x + 6x − 5 = 3
set it equal to the asymptote value
3x + 6 = 3x − 15
multiply by (x − 5)
6 = −15
impossible
No x works, so the curve never touches y = 3 — that is what makes it an asymptote. (A horizontal asymptote can be crossed in other functions; this test tells us when.)
A hole and a wall together: f(x) = x − 1x² + 2x − 3
= x − 1(x − 1)(x + 3)
factor the bottom
= 1x + 3, x ≠ 1
(x − 1) cancels
hole at (1, 14), wall at x = −3
11 + 3 = 14
Only the factor that cancels makes a hole; the factor left on the bottom still makes a wall.
The leading term written last: f(x) = 3 − 6x2x + 7
top leading coefficient −6
the x-term, not the first number
bottom leading coefficient 2
degrees tie
y = −62 = −3
a negative asymptote
Reading 32 from the first numbers is the slip. The leading coefficient belongs to the highest power of x, wherever it is written.
No y-intercept: f(x) = x + 2x
f(0) = 20
undefined
no y-intercept; wall at x = 0
the y-axis is the asymptote
zero at x = −2
from the top
When 0 is excluded from the domain, the graph cannot cross the y-axis at all.
Periodical exams ask for one feature at a time — the zero, the y-intercept, the vertical or horizontal asymptote, or a hole — and often a sketch from all four. Entrance tests like the hole that looks like a wall and the "which function has these features" question. Factor top and bottom first, cancel, and read each feature from its own place. For a 10-second check, put the claimed zero into the original: the top must be 0 and the bottom must not.
Find the horizontal asymptote
f(x) = 2x + 1x − 4
Both top and bottom have degree 1, so the horizontal asymptote is the ratio of the leading coefficients: 21, giving y = 2. A is the rule for when the top degree is smaller, which is not the case here. B is the vertical asymptote. D divides the constants instead of the leading coefficients.
Hint: look at the highest power on each line, not at the numbers on their own. Both are x to the first power, so the degrees tie.
