SME
Grade 9 Math · Algebra 1
Piecewise, step and absolute value functions
Use this for “f(x) is x + 2 for x < 1 and −2x + 7 for x ≥ 1. Find f(3).”
Chart
f(3) = 1
Left piece’s end at the break: 3
Value at the break, f(p): 5
This is an example. Type your own values to start a new problem.
f(x)
=
m₁
x
+
b₁
for
x
<
p
f(x)
=
m₂
x
+
b₂
for
x
≥
p
f(
)
=
y
L
Left piece’s end at the break
worked out (not typed)
yₚ
Value at the break, f(p)
worked out (not typed)
Formulas
y
=
m₁
×
x
+
b₁
if
x
<
p
,
m₂
×
x
+
b₂
if
x
≥
p
1 = 1 × 3 + 2 if 3 < 1, −2 × 3 + 7 if 3 ≥ 1
L
=
m₁
×
p
+
b₁
3 = 1 × 1 + 2
yₚ
=
m₂
×
p
+
b₂
5 = −2 × 1 + 7
Assumptions
•
Use the piece whose condition x meets.
•
At the break, the closed dot is the value; the open dot is only where the other piece heads.
•
The pieces need not meet.
Step-by-step
WE KNOW
x = 3, p = 1, m₁ = 1, b₁ = 2, m₂ = −2, b₂ = 7
FIND
output f(x) (y), left piece’s end at the break (L), value at the break, f(p) (yₚ)
Step 1 · Find output f(x): y
Use y = m₁ × x + b₁ if x < p, m₂ × x + b₂ if x ≥ p
x
is
at
the
break
or
right
of
it
(
x
≥
p
):
use
the
right
piece.
y
=
m₂
×
x
+
b₂
y
=
−2
×
3
+
7
y
=
−6
+
7
y
=
1
Step 2 · Find left piece’s end at the break (L)
Use L = m₁ × p + b₁
The
left
piece
heads
here
at
the
break,
but
x
<
p
leaves
it
out:
an
open
dot.
L
=
1
×
1
+
2
L
=
3
Step 3 · Find value at the break, f(p): yₚ
Use yₚ = m₂ × p + b₂
x
≥
p
takes
the
break
into
the
right
piece:
the
closed
dot
is
f
(
p
).
yₚ = −2 × 1 + 7
yₚ = 5
Check
Put the numbers back in. Both sides should match.
1 = −2 × 3 + 7 ✓
3 = 1 × 1 + 2 ✓
5 = −2 × 1 + 7 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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