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2010
P1 Q5
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10 P1 Q5
2010 H2 Mathematics Paper 1 Question 5
Graphs and Transformations
Answers
(i)
y
=
1
2
(
x
−
2
)
3
−
6
{y = \frac{1}{2}(x-2)^3 - 6}
y
=
2
1
(
x
−
2
)
3
−
6
(
2
+
12
3
,
0
)
{\left( 2+\sqrt[3]{12}, 0 \right)}
(
2
+
3
12
,
0
)
(
0
,
−
10
)
{\left( 0, -10 \right)}
(
0
,
−
10
)
(ii)
(
−
10
,
0
)
{\left( -10, 0 \right)}
(
−
10
,
0
)
(
0
,
2
+
12
3
)
{\left( 0, 2+\sqrt[3]{12} \right)}
(
0
,
2
+
3
12
)
Full solutions
(i)
y
=
x
3
↓
y
=
(
x
−
2
)
3
↓
y
=
1
2
(
x
−
2
)
3
↓
y
=
1
2
(
x
−
2
)
3
−
6
■
\begin{gather*} y = x^3 \\ \downarrow \\ y = (x-2)^3 \\ \downarrow \\ y = \frac{1}{2}(x-2)^3 \\ \downarrow \\ y = \frac{1}{2}(x-2)^3 - 6 \; \blacksquare \end{gather*}
y
=
x
3
↓
y
=
(
x
−
2
)
3
↓
y
=
2
1
(
x
−
2
)
3
↓
y
=
2
1
(
x
−
2
)
3
−
6
■
When the curve crosses the
x
-
{x\textrm{-}}
x
-
axis,
y
=
0
{y=0}
y
=
0
1
2
(
x
−
2
)
3
−
6
=
0
(
x
−
2
)
3
=
12
x
=
2
+
12
3
\begin{gather*} \frac{1}{2}(x-2)^3 - 6 = 0 \\ (x-2)^3 = 12 \\ x = 2 + \sqrt[3]{12} \\ \end{gather*}
2
1
(
x
−
2
)
3
−
6
=
0
(
x
−
2
)
3
=
12
x
=
2
+
3
12
When the curve crosses the
x
-
{x\textrm{-}}
x
-
axis,
x
=
0
{x=0}
x
=
0
y
=
1
2
(
0
−
2
)
3
−
6
=
−
10
\begin{align*} y &= \frac{1}{2}(0-2)^3 - 6 \\ &= -10 \end{align*}
y
=
2
1
(
0
−
2
)
3
−
6
=
−
10
Coordinates of points where the curve crosses the
x
-
{x\textrm{-}}
x
-
and
y
-
{y\textrm{-}}
y
-
axes:
(
2
+
12
3
,
0
)
■
(
0
,
−
10
)
■
\begin{align*} &\left( 2+\sqrt[3]{12}, 0 \right) \; \blacksquare \\ &\left( 0, -10 \right) \; \blacksquare \\ \end{align*}
(
2
+
3
12
,
0
)
■
(
0
,
−
10
)
■
(ii)
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