SME
Grade 9 Math · Problem type
Absolute value function
Use this for “Graph y = −2|x − 1| + 6: its vertex, which way it opens, and its zeros.”
Chart
Vertex (1, 6), axis of symmetry x = 1
Zeros: x = −2, x = 4
f(0) = 4
This is an example. Type your own values to start a new problem.
y
=
a
|x
−
h
|
+
k
d
Distance from h to each zero
worked out (not typed)
x₁
Left zero
worked out (not typed)
x₂
Right zero
worked out (not typed)
x
Input
example
y
Output
calculated
Formulas
y
=
a
×
|
x
−
h
|
+
k
4 = −2 × |0 − 1| + 6
d
=
−
k
a
3
=
−
6
−2
x₁
=
h
−
d
−2 = 1 − 3
x₂
=
h
+
d
4 = 1 + 3
Assumptions
•
y = a|x − h| + k is a V with its vertex at (h, k).
•
a > 0 opens up, a < 0 opens down; a bigger |a| is steeper.
•
The zeros solve a|x − h| + k = 0: |x − h| = −k ÷ a.
Step-by-step
WE KNOW
x = 0, a = −2, h = 1, k = 6
FIND
output (y), distance from h to each zero (d), left zero (x₁), right zero (x₂)
Step 1 · Find output (y)
Use y = a × |x − h| + k
Take
the
distance
from
x
to
h
,
multiply
by
a
,
then
add
k
.
y
=
−2
×
|0
−
1|
+
6
y
=
−2
×
1
+
6
y
=
−2
+
6
y
=
4
Step 2 · Find distance from h to each zero (d)
Use d = −k ÷ a
Set
y
=
0:
a
|
x
−
h
|
=
−
k
,
so
|
x
−
h
|
=
−
k
a
:
each
zero
is
this
far
from
h
.
d
=
−
6
−2
d
=
3
Step 3 · Find left zero (x₁)
Use x₁ = h − d
The
left
zero
is
d
left
of
h
.
x₁
=
1
−
3
x₁
=
−2
Step 4 · Find right zero (x₂)
Use x₂ = h + d
And
the
right
zero
is
d
right
of
h
.
x₂
=
1
+
3
x₂
=
4
Check
Put the numbers back in. Both sides should match.
4 = −2 × |0 − 1| + 6 ✓
3
=
−
6
−2
✓
−2 = 1 − 3 ✓
4 = 1 + 3 ✓
Topic check
See what you know: 5 questions on this topic
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