SME
Grade 11 Math · Algebra 2
Rational expressions, equations and functions
Use this for “Find the vertical asymptote and the hole of y = (x − 1)(x + 2)/((x − 3)(x + 2)).”
Chart
Zero: x = 1
Vertical asymptote x = 3
Horizontal asymptote y = 1
Hole at (−2, 0.6)
f(5) = 2
y of the hole: 3/5
This is an example. Type your own values to start a new problem.
y
=
(x
−
)(x
−
(
)
)
(x
−
)(x
−
(
)
)
H
y of the hole
worked out (not typed)
x
Input
example
y
Output
worked out (not typed)
Formulas
H
=
a
×
c
−
z
c
−
p
3
5
=
1
×
−2 − 1
−2 − 3
y
=
a
×
x
−
z
x
−
p
2
=
1
×
5 − 1
5 − 3
Assumptions
•
y = a(x − z)(x − c) ÷ ((x − p)(x − c)): a factor that cancels leaves a hole at x = c, not an asymptote.
•
A factor left in the denominator gives a vertical asymptote, x = p.
•
Equal degrees top and bottom: the horizontal asymptote is y = a.
Step-by-step
WE KNOW
a = 1, z = 1, p = 3, c = −2, x = 5
FIND
y of the hole (H), output (y)
Step 1 · Find y of the hole (H)
Use H = a × (c − z) ÷ (c − p)
Cancel
x
−
c
,
then
put
x
=
c
into
what
is
left:
the
hole’s
height.
H
=
1
×
−2 − 1
−2 − 3
H
=
−
3
−5
H
=
3
5
Step 2 · Find output (y)
Use y = a × (x − z) ÷ (x − p)
Cancel
the
shared
factor
x
−
c
,
then
evaluate.
y
=
1
×
5 − 1
5 − 3
y
=
4
2
y
=
2
Check
Put the numbers back in. Both sides should match.
3
5
=
1
×
−2 − 1
−2 − 3
✓
2
=
1
×
5 − 1
5 − 3
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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