Mathematics · May / June 2021 · Paper 11 · Question 8
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The question as text
8. The curve C1 has equation y = 3x2 + 6x + 9
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8(a)[3]
3
x
2
+
6
x
+
9
=
3(
x
±
...)
2
±
...
a =
3
&
b =
3
x
2
+
6
x
+
9
=
3(
x
+
1)
2
±
...
a =
3
1
3
x
2
+
6
x
+
9
=
3(
x
+
1)
2
+
6
8(b)[1]
( 1, 6)
−
8(c)[5]
y
=
α
(
x
+
4)(
x
+
2)(
x
−
3)
6
=
α
( 1
−+
4)( 1
−+
2)( 1 3)
−−
1
α = −
2
1
y
= −
"
"(
x
+
4)(
x
+
2)(
x
−
3)
⇒
y
=
...
x
3
+
...
x
2
+
...
x
+
...
2
1
3
A
= −
,
B
= −
,
C
=
5,
D
=
12
2
2
Alt (c)
−
64
A
+
16
B
−
4
C
+
D
=
0
−
8
A
+
4
B
−
2
C
+
D
=
0
27
9
3
0
A
+
B
+
C
+
D
=
−
A
+
B
−
C
+
D
=
6
1
3
One of
A
= −
,
B
= −
,
C
=
5,
D
=
12
2
2
Fully solves their simultaneous equations
1
3
A
= −
,
B
= −
,
C
=
5,
D
=
12
2
2
(a)
B1
Achieves
3
x
2
+
6
x
+
9
=
3(
x
±
...)
2
±
...
or states that
a =
3
M1
Deals correctly with the first two terms of
3
x
2
+
6
x
+
9
Scored for
3
x
2
+
6
x
+
9
=
3(
x
+
1)
2
±
...
or states that
a =
3
&
b =
1
A1
3
x
2
+
6
x
+
9
=
3(
x
+
1)
2
+
6
This may be done by equating coefficients using the expanded form
a x
(
+
b
)
2
+
c
=
ax
2
+
2
abx
+
ab
2
+
c
(b)
B1ft
( 1,6)
−
or follow through their (
−
b c
, )
from (a). Condone lack of brackets and accept
eg
x
= −
1,
y
=
6
(c)
B1
Identifies that three factors of the cubic equation are (
x
+
4)(
x
+
2)(
x
−
3)
M1
A correct method to find the scale factor by using the minimum point found in part (c). Look
for the minimum point to be substituted into their equation for the cubic
1
A1
Scale factor =
−
2
M1
Attempts to multiply (
x
±
4)(
x
±
2)(
x
±
3)
to achieve
x +
3
........
±
24
. This may have been
multiplied by their scale factor so look for
α
(
x
3
+
........
±
24)
1
3
1
3
A1
A
,
B
,
C
5,
D
12
. Accept
y
x
3
x
2
5
x
12
= −
= −
=
=
= −
−
+
+
2
2
2
2
In the alternative method using simultaneous equations
B1
Three correct equations formed using x = 4,
−
−
2 and 3
with
y = in each case.
0
M1
Forms four simultaneous equations using the three intercepts and their point P
1
3
A1
One of
A
= −
,
B
= −
,
C
=
5,
D
=
12
2
2
M1
Fully solves their simultaneous equations either using matrices or elimination. This may be
done on a calculator which is acceptable.
−
1
−
64
16
−
4
1
0
A
A
−
8
4
−
2
1
0
B
B
Eg. Using matrices
=
⇒
=
...
27
9
3
1
0
C
C
−
1
1
−
1 1
6
D
D
1
3
1
3
,
,
5,
12
3
2
5
12
A1
A
= −
B
= −
C
=
D
=
. Accept
y
= −
x
−
x
+
x
+
2
2
2
2
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