Draw the graph of y x 2 2
Calculate points from this table and plot the points as you go.
y
y x2 2
( x, y )
2
y (2) 2 2 6
1
y (1) 2 2 3
(2,6)
(1,3)
0
y 02 2 2
(0,2)
1
y 12 2 3
(1,3)
2
y 22 2 6
(2,6)
x
x
Draw
a parabola
through
theon
points.
The basic
parabola
is shown
the graph, dotted blue.
Adding 2 to x2 moves the basic parabola
How?
up by
2 units.
Draw the graph of y x 2 3
Calculate points from this table and plot.
x
2
1
y x2 3
y
( x, y )
y (2) 2 3 1
(2,1)
y (1) 2 3 2 (1,2)
0
y 02 3 3
(0,3)
1
y 12 3 2
(1,2)
2
y 22 3 1
x
(2,1)
The
basic
parabolathrough
is shown
the graph, dotted blue.
Draw
a parabola
theonpoints.
Subtracting 3 from x2 moves the basic parabola
How?by 3 units.
down
Draw the graph of y ( x 4) 2
y
Calculate points from this table
and plot.
x
y ( x 4) 2
( x, y )
5
y (5 4) 2 1
4
y (4 4) 2 0
(5,1)
(4,0)
3
y (3 4) 2 1
(3,1)
2
y (2 4) 2 4
y (1 4) 2 9
(2,4)
1
(1,9)
Draw
a parabola
through
theon
points.
The
basic
parabola
is shown
the graph, dotted blue.
Adding 4 to x and then squaring moves the basic parabola
to the left by 4 units.
How?
x
Draw the graph of y ( x 1) 2
y
Calculate points from this table
and plot.
y ( x 1) 2
( x, y )
1
y (1 1) 2 4
(1,4)
0
y (0 1) 2 1
(0,1)
1
y (1 1) 2 0
(1,0)
2
y (2 1) 2 1
(2,1)
3
y (3 1) 4
x
2
(3,4)
The basic
parabola
is shown
the graph, dotted blue.
Draw
a parabola
through
theon
points.
Subtracting 1 from x and then squaring moves the basic parabola
How?
to the right by 1 unit.
x
Draw the parabola given by y ( x 2) 2 3
y
y ( x 2) 2 3
The
2 moves the parabola
2 units to the right.
How?
x
Draw the parabola given by y ( x 2) 2 3
y
y ( x 2) 2 3
The
3 moves the parabola
3 units down.
How?
x
Draw the graph of y 2 x 2
y
Calculate points from this table
and plot.
x
2
y 2x 2
( x, y )
(2,8)
y 2 (2) 2 8
y 2 (1) 2 2
(1,2)
0
y 2 02 0
(0,0)
1
y 212 2
(1,2)
2
y 2 2 8
1
2
x
(2,8)
The
basic
parabola
is shown
the graph, dotted blue.
Draw
a parabola
through
theon
points.
Multiplying x2 by 2 makes the basic parabola
steeper.
What?
1
Draw the graph of y x 2
4
y
Calculate points from this table.
x
4
2
0
2
4
1 2
y x
4
1 2
y ( 4) 4
4
1
y (2) 2 1
4
1 2
y 0 0
4
1
y 22 1
4
1 2
y 4 4
4
( x, y )
(4,4)
(2,1)
(0,0)
(2,1)
x
(4,4)
Draw
a parabola
through
theon
points.
The basic
parabola
is shown
the graph, dotted blue
Multiplying x2 by 1 makes the basic parabola
4
wider.
What?
Draw the parabola given by y x 2 4
y
y x2 4
The x2 means the parabola is
upside down.
What?
x
Draw the parabola given by y x 2 4
y x2 4
y
The 4 means the parabola is moved
up by 4 units.
How?
x
Draw the parabola given by y ( x 2) 2
y
y ( x 2) 2
The
–(….)2 means the parabola is
upside down.
What?
x
Draw the parabola given by y ( x 2) 2
y
y ( x 2) 2
The 2 means the parabola is moved
left by 2 units.
How?
x
Draw the parabola given by y ( x 1) 2 2
y
y ( x 1) 2 2
The
–(….)2 means the parabola is
upside down.
What?
x
Draw the parabola given by y ( x 1) 2 2
y ( x 1) 2 2
–
The 1 means the parabola is moved
right
How?by 1 unit.
x
Draw the parabola given by y ( x 1) 2 2
y ( x 1) 2 2
–
The 2 means the parabola is moved
How?
down by 2 units.
x
Draw the graph of y ( x 3)( x 1)
y
Calculate points from this table & plot.
y ( x 3)( x 1)
( x, y )
2 y (2 3)(2 1) 5 (2,5)
1 y (1 3)(1 1) 0 (1,0)
0 y (0 3)(0 1) 3 (0,3)
x
1 y (1 3)(1 1) 4
2 y (2 3)(2 1) 3
(1,4)
(2,3)
3 y (3 3)(3 1) 0
(3,0)
4 y (4 3)(4 1) 5
(4,5)
–
Draw
a parabola
the points.
The graph
showsthrough
the parabola
cuts the x-axis at 3 and 1.
Continued on next slide.
x
Draw the graph of y ( x 3)( x 1)
The x-intercepts can be obtained from the
two factors:
y
y ( x 3)( x 1)
At x-intercepts, y = 0.
( x 3)( x 1) 0
Solve.
x 3 0 or x 1 0
x 3 or x 1
Coordinates of the x-intercepts are (3,0) and (1,0).
x
Sketch y x 2 2 x 8
Factorise
x 2 x 8 ( x 4)( x 2)
2
y
Equate to 0 for x-intercepts
Solve.
( x 4)( x 2) 0
x 4 or x 2
The x - intercepts are (4,0) and (2,0)
Substitute x = 0 for y-intercept
y 0 2 0 8 8
The y - intercept is (0,8)
2
The line of symmetry of the parabola
is halfway between x = 4 and x = 2.
i.e. x = 1.
Substitute x = 1 for y-coordinate of vertex
y 12 2 1 8 1 2 8 9
(1,9)
The vertex is at
Draw a parabola through the points.
x
Sketch y x( x 4)
y
Equate to 0 for x-intercepts
Solve.
x 0 or x 4
x( x 4) 0
The x - intercepts are (0,0) and (4,0)
Substitute x = 0 for y-intercept
y 0 (0 4) 0
Draw a parabola through the points.
The y - intercept is (0,0)
The line of symmetry of the parabola
is halfway between x = 0 and x = 4.
x
i.e. x = 2.
Substitute x = 2 for y-coordinate of vertex
y 2 (2 4) 2 2 4
The vertex is at
(2,4)
Sketch y (1 x)( x 5)
y
Equate to 0 for x-intercepts
(1 x)( x 5) 0 Solve.
1 x 0 or x 5 0
x 1 or x 5
The x - intercepts are (1,0) and (5,0)
Substitute x = 0 for y-intercept
y (1 0) (0 5) 5
The y - intercept is (0,5)
Note: The position of the vertex
means the graph must be upside
down.
Draw a parabola through the points.
The line of symmetry of the parabola
is halfway between x = 1 and x = 5.
x
i.e. x = 3.
Substitute x = 3 for y-coordinate of vertex
y (1 3) (3 5) 2 2 4
The vertex is at
(3,4)
0
