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  212  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)