SME
Grade 9 Math · Algebra 1
Graphing quadratic functions: vertex, axis of symmetry and zeros
Use this for “Graph y = 2(x − 1)² − 8: its vertex, axis of symmetry, zeros and y-intercept.”
Chart
Vertex (1, −8), axis of symmetry x = 1
Zeros: x = −1, x = 3
y-intercept (0, −6)
Range: [−8, ∞)
f(4) = 10
This is an example. Type your own values to start a new problem.
y
=
a
(x
−
h
)²
+
(
k
)
c
y-intercept
calculated
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
)
2
+
k
10
=
2
×
(4 − 1)
2
+
(−8)
c
=
a
×
h
2
+
k
−6
=
2
×
1
2
+
(−8)
x₁
=
h
−
√(−
k
a
)
−1
=
1
−
√(−
−8
2
)
x₂
=
h
+
√(−
k
a
)
3
=
1
+
√(−
−8
2
)
Assumptions
•
The vertex is (h, k) and the axis of symmetry is x = h.
•
a > 0 opens up, with a least value k; a < 0 opens down, with a greatest value k.
•
The zeros are where y = 0; the y-intercept is where x = 0.
Step-by-step
WE KNOW
x = 4, a = 2, h = 1, k = −8
FIND
output (y), y-intercept (c), left zero (x₁), right zero (x₂)
Step 1 · Find output (y)
Use y = a × (x − h)² + k
Find
the
distance
from
h
,
square
it,
multiply
by
a
,
then
add
k
.
y
=
2
×
(4 − 1)
2
−
8
y
=
2
×
3
2
−
8
y
=
2
×
9
−
8
y
=
18
−
8
y
=
10
Step 2 · Find y-intercept (c)
Use c = a × h² + k
Put
x
=
0
into
the
rule:
y
=
a
(0 −
h
)
2
+
k
=
ah²
+
k
.
c
=
2
×
1
2
−
8
c
=
2
−
8
c
=
−6
Step 3 · Find left zero (x₁)
Use x₁ = h − √(−k ÷ a)
Set
y
=
0:
a
(
x
−
h
)
2
=
−
k
,
so
x
−
h
=
±√(−
k
a
).
The
minus
gives
the
left
zero.
x₁
=
1
−
√(−
−8
2
)
x₁
=
1
−
√(
8
2
)
x₁
=
1
−
√
4
x₁
=
1
−
2
x₁
=
−1
Step 4 · Find right zero (x₂)
Use x₂ = h + √(−k ÷ a)
The plus gives the right zero, as far right of the axis as the left one is left of it.
x₂
=
1
+
√(−
−8
2
)
x₂
=
1
+
√(
8
2
)
x₂
=
1
+
√
4
x₂
=
1
+
2
x₂
=
3
Check
Put the numbers back in. Both sides should match.
10
=
2
×
(4 − 1)
2
−
8
✓
−6
=
2
×
1
2
−
8
✓
−1
=
1
−
√(−
−8
2
)
✓
3
=
1
+
√(−
−8
2
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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