SME
Grade 9 Math · Problem type
Graph of |f(x)|
Use this for “Graph y = |x² − 2x − 3|” from the graph of f.
Chart
Zeros: x = −1, x = 3
f(1) = 4
f(x) = |x² − 2x − 3|: the parts below the x-axis are reflected up; the curve before is dashed
This is an example. Type your own values to start a new problem.
y
=
|
a
x²
+
(
b
)
x
+
(
c
)
|
x
Input
example
u
f(x) before the bars
worked out (not typed)
y
Output |f(x)|
worked out (not typed)
Formulas
u
=
a
×
x
2
+
b
×
x
+
c
−4
=
1
×
1
2
+
(−2)
×
1
+
(−3)
y
=
|
u
|
4 = |−4|
Assumptions
•
|f(x)| is f(x) where f(x) ≥ 0 and −f(x) where f(x) < 0.
•
So the parts of the graph below the x-axis flip up over it; the rest stays.
•
The zeros of f stay put: there the graph touches the x-axis and turns sharply.
Step-by-step
WE KNOW
a = 1, b = −2, c = −3, x = 1
FIND
f(x) before the bars (u), output |f(x)| (y)
Step 1 · Find f(x) before the bars (u)
Use u = a × x² + b × x + c
Put
x
into
f
first.
u
=
1
×
1
2
−
2
×
1
−
3
u
=
1
−
2
−
3
u
=
−4
Step 2 · Find output |f(x)| (y)
Use y = |u|
Then
take
its
absolute
value:
a
negative
f
(
x
)
turns
positive.
y
=
|−4|
y
=
4
Check
Put the numbers back in. Both sides should match.
−4
=
1
×
1
2
−
2
×
1
−
3
✓
4 = |−4| ✓
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