The denominator decides the walls, the degrees decide the ceiling, and a factor that cancels leaves a hole instead.
In the xy-plane, the graph of y = f(x) has a vertical asymptote at x = a and a horizontal asymptote at y = b. What is the value of a + b?
f(x) =
4x + 9x − 5
The denominator is zero at x = 5 and the numerator is not, so a = 5. Top and bottom are both degree 1, so the graph flattens out at the ratio of the leading coefficients, b = 41 = 4. Then a + b = 9. A solves x + 5 = 0 and gets −5 for the wall; B assumes the horizontal asymptote is always y = 0, which is only true when the top has the lower degree; D reads the 9 in the numerator as the horizontal asymptote instead of comparing leading coefficients.
1
Start where the function is undefined.
A fraction dies when its denominator is zero. Set the denominator equal to zero and solve — that single x is the only candidate for a vertical asymptote.
x − 5 = 0 → x = 5
f(5) does not exist
f(4.9) = 28.6 ÷ (−0.1) = −286
a hair to the left, the value plunges
f(5.1) = 29.4 ÷ 0.1 = 294
a hair to the right, it soars
vertical asymptote: x = 5
That is all a vertical asymptote is: the numerator stays near a normal
number while the denominator shrinks to nothing, so the quotient runs
off the screen. The curve never touches the line x = 5 — it is a wall,
not a point on the graph.
2
The horizontal asymptote is a comparison of degrees.
Far out to the left and right, only the highest power of x in each part still matters.
Degree of top vs bottom
Horizontal asymptote
Example
top lower
y = 0
3x + 1x² + 2
equal
y = ratio of leading coefficients
4x + 9x − 5 → y = 4
top higher
none — the values escape
x² + 1x − 5
f(1,000) = 4,009 ÷ 995 ≈ 4.029
closing in on 4 from above
f(−1,000) = (−3,991) ÷ (−1,005) ≈ 3.971
closing in on 4 from below
horizontal asymptote: y = 4
y = 0 is a reflex, not an answer
The choice 5 in the opening problem is what you get by taking the
horizontal asymptote to be y = 0 without looking at the degrees, and 14
comes from grabbing the 9 out of the numerator. Compare the degrees
first; only then read a number.
3
Factor first: a factor that cancels leaves a hole, not a wall.
Survives the cancelling → vertical asymptote. Cancels away → a single missing point.
k(x) = x² − x − 6x − 3
the denominator is zero at x = 3
= (x − 3)(x + 2)x − 3
the numerator is zero at x = 3 too
= x + 2, for every x except 3
a line with one point punched out
the hole sits at (3, 5)
put x = 3 into the cancelled form
A hole is invisible on a calculator screen
Graph k in Desmos and you see a plain straight line — the missing point
is one pixel wide. That is exactly why the SAT asks about it: only the
factoring shows you that x = 3 is not in the domain. Factor before you
trust a picture.
The SAT draws them for you as dashed lines. The vertical one is an x-value; the horizontal one is a y-value.
the dashed upright line sits at x = −2
so the denominator is x + 2
the dashed flat line sits at y = 1
the value the two branches flatten toward
y = x + 5x + 2 = 1 + 3x + 2
equal degrees, ratio 11 = 1 ✓
Notice what the two branches do. Each one runs away along the wall and
settles along the ceiling, and the ceiling is approached from above on
one side and from below on the other. A curve may cross its horizontal
asymptote somewhere in the middle; it can never cross a vertical one.
5
Know which question the SAT is actually asking.
Four wordings, one piece of work: factor, then read.
The stem says
You look at
“for what value of x is f undefined”
every zero of the denominator, cancelled or not
“an equation of the vertical asymptote”
the denominator zeros that survive cancelling
“the graph approaches y = …”
the degree comparison
“the graph crosses the x-axis at”
the numerator zeros that survive cancelling
The last row is the one that gets swapped. Numerator zeros are where the
graph quietly passes through zero; denominator zeros are where it leaves
the page. They are opposite ends of the same fraction.
Use Desmos to confirm, never to decide
Type the function in and the walls show up as near-vertical streaks; slide
the window out to x = 50 and the curve flattens at its horizontal
asymptote, so you can read the number off the y-axis. What the screen will
not tell you is a hole, or whether a steep streak is at x = 4.9 or x = 5.
Do the factoring; let the graph confirm it.
✓
Check: test a big x and a nearby x.
A horizontal asymptote is confirmed far out; a vertical one is confirmed up close.
f(1,000)4,009 ÷ 995 ≈ 4.029the ceiling really is y = 4 ✓
f(4.9)28.6 ÷ (−0.1) = −286the wall really is at x = 5 ✓
k near the holek(2.99) = 4.99, k(3.01) = 5.01heads for 5, never arrives ✓
the graphed curveat x = 1: 1 + 51 + 2 = 2matches the point on the right branch ✓
Remember the flow
Factor both · Cancelled zero = hole · Surviving zero = vertical · Degrees = horizontal
The curve may cross the ceiling. It never crosses the wall.
Now you try
Practice problem
In the xy-plane, the graph of y = g(x) has exactly one vertical asymptote, at x = a, and exactly one hole, at x = b. What is the value of a + b?
g(x) =
x² + 3x − 10x² + x − 6
Factor both: g(x) = (x + 5)x − 2(x + 3)(x − 2). The factor x − 2 cancels, so x = 2 is a hole and b = 2; the surviving denominator zero is x = −3, so a = −3. Then a + b = −1. A takes x = −5 as the hole — that numerator zero does not cancel, so it is an x-intercept; B slips the sign on the cancelled factor, reading x − 2 = 0 as x = −2; D slips the sign on x + 3 and puts the wall at x = 3. Sanity check: away from the hole g(x) = x + 5x + 3, and g(100) = 105103 ≈ 1.019, so the horizontal asymptote is y = 1.
Hint: factor the top and the bottom before you read a single
zero. Then sort the denominator's zeros into two piles — the one that also
appears on top, and the one that does not.
