Translations and
Completing the Square
© Christine Crisp
Translations
The graph of
parabola
yx
2
forms a curve called a
y  x2
This point . . . is called the vertex
Translations
 Adding a constant translates y  x up the y-axis

y  x2
y  x2  3
e.g.
2
y  x2
y  x2  3

y  x2
The vertex is now ( 0, 3)
y x
2
 y x 3
2
has added 3 to the y-values
Translations
Adding 3 to x gives
y  x2
We get

y  ( x  3) 2
y  x2
y  ( x  3)
2
Adding 3 to
x moves the
curve 3 to
the left.
This may seem surprising but on the x-axis, y = 0
so,
y0 
0  ( x  3) 2

x 3
Translations
 Translating in both directions
2
2
y

x
y

(x

5
)
3
e.g.

y  (x  5) 2  3
y  x2
We can write this in vector form as:
translation 5
3
 
Translations
SUMMARY
 The curve y  ( x  p)  q
2
is a translation of y  x
2
by
 The vertex is given by ( p, q)
  p
 q 
 
Translations
Exercises: Sketch the following translations of y  x 2
2
2

y

x
y

(x

2
)
1
1.
y  x2
y  ( x  2)2  1
2. y  x
3.
2
 y  (x  3)  2
y  ( x  3)2  2
2
y  x2
y  x 2  y  (x  4) 2  3
y  x2
y  ( x  4)2  3
Translations
y  x2
Sketch the following translations of
2
2

y

x
y

(x

2
)
1
1.
y  x2
y  ( x  2)2  1
Now insert a coefficient infront of the bracket
y  x 2 y  2( x  2)  1
2
5 ^|
Y
4
y  2( x  2)2  1
3
y  x2
2
The 2 outside the bracket
has stretched the curve
vertically by a factor of 2
y  ( x  2)  1
1
2
-4
-3
-2
-1
0
-1
-2
1
2
X->
3
Translations
4 Sketch the curve found by translating
y  x by
2
 2
  3 .
 
What is its equation?
y  (x  2) 2  3
5 Sketch the curve found by translating
y x
2
 1
by   .
2
What is its equation?
y  (x  1) 2  2
Translations and Completing the Square
A quadratic function which is written in the
form
y  ( x  p) 2  q
is said to be in its completed square form.
We often multiply out the brackets as follows:
e.g.
y  (x  5) 2  3
This means multiply
( x – 5 ) by 
itself y  ( x  5)( x  5)  3
So

y  x 2  5 x  5 x  25  3

y  x 2  10x  28
y  (x  5)  3  x  10 x  28
2
2
Completing the Square
The completed square form of a quadratic
function
• writes the equation so we can see the
2
translation from y  x
• gives the vertex
Completing the Square
e.g. Consider y  x translated by 2 to
the left and 3 up.
We can write this in
vector form as:
 2
translation  
3
The equation of the curve is
2
y  (x  2)  3
2
Completed square form
Check: The vertex is ( -2, 3)
Completing the Square
Any quadratic expression which has the form
ax2 + bx + c can be written as p(x + q)2 + r
2x2 + 4x + 5 = 2(x + 1)2 + 3
This can be checked by multiplying out the bracket
2(x + 1)2 + 3 = 2(x + 1)(x + 1) + 3
= 2(x2 + x + x + 1) + 3
= 2x2 + 4x + 2 + 3
= 2x2 + 4x + 5
Completing the Square
2x2 + 4x + 5 = p(x + q)2 + r = 2(x + 1)2 + 3
We need to find the values of p, q and r
If the graph is plotted we find that the vertex is at (–1, 3)
The graph has a Horizontal translation of –1 so q = +1
Opposite sign
Vertical translation of 3
Same Sign
so r = 3
p = coefficient of x2
p=2
So 2x2 + 4x + 5 = p(x + q)2 + r = 2(x + 1)2 + 3
Completing the Square
Using a Calculator to Complete the Square
1) Plot the given curve in y1
2) Use the window button to set the scale so that the vertex
is clearly visible
3) Use the 2ndF Trace button (Calc) to find the vertex –
either a maximum or a minimum
4) Fill in the horizontal translation q
5) Fill in the vertical translation r
6) Fill in the vertical stretch using the coefficient of x2
Completing the Square
Ex1 y = 2x2 – 3x – 5
Express in the form p(x + q)2 + r
5 ^|
Y
4
Y=2x^2-3x-5
3
Using 2ndF Trace button (Calc) to find the
vertex
Min at x = 0.75 and y = –6.125
Horizontal translation of + 0.75 so q = –0.75
Opposite sign
2
1
-2
-1
0
-1
-2
-3
-4
-5
-6
-7
Vertical translation of –6.125 so r = –6.125
Same Sign
p = coefficient of x2 = 2
So 2x2 – 3x – 5 = 2(x – 0.75)2 – 6.125
-8
-9
1
2
3
X->
4
Completing the Square
Ex1 y = 4 – 3x – x2
Express in the form p(x + q)2 + r
9 ^|
Y
8
Y=4-3x-x^2
Using 2ndF Trace button (Calc) to find the
vertex
Max at x = –1.5 and y = 6.25
7
6
5
4
3
Horizontal translation of –1.5 so q = 1.5
Opposite sign
2
1
-4
-3
-2
-1
0
-1
Vertical translation of 6.25
Same Sign
so r = 6.25
p = coefficient of x2 = –1
So 4 – 3x – x2 = –1(x + 1.5)2 + 6.25
-2
-3
1
2
3
X->
4
Completing the Square
Exercises
Complete the square for the following quadratics:
1.
x 2  4x  6
 (x  2) 2  2
2.
x  4x  3
 (x  2) 2  7
3.
x 2  6 x  10
 (x  3) 2  1
2
Completing the Square
4.
x  8x  2
 (x  4) 2  18
5.
x 2  3x  3
 (x  32 ) 2  43
6.
2x 2  8x  1
 2(x  2) 2  7
2
Qu.s in notes pg 32