Translations and Completing the Square © Christine Crisp.
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Transcript of Translations and Completing the Square © Christine Crisp.
Translations
2xy
The graph of forms a curve called a parabola
2xy
This point . . . is called the vertex
Translations
32 xy2xy
2xy
Adding a constant translates up the y-axis
2xy 32 xye.g.
2xy
The vertex is now ( 0, 3)
has added 3 to the y-values
2xy 32 xy
Translations
This may seem surprising but on the x-axis, y = 0so, x 3
We get
230 )( x0y
Adding 3 to x gives 23)( xy2xy
Adding 3 to x moves the curve 3 to the left.
23)( xy
2xy
Translations
Translating in both directions 35 2 )(xy2xy e.g.
3
5
We can write this in vector form as:
translation
35 2 )(xy2xy
Translations
Exercises: Sketch the following translations of 2xy
12 2 )(xy2xy 1.
23 2 )(xy2xy 2.
34 2 )(xy2xy 3.
1)2( 2 xy
2xy
2xy
2)3( 2 xy
2xy
3)4( 2 xy
Translations
Sketch the following translations of2xy
12 2 )(xy2xy 1.
1)2( 2 xy
2xy
Now insert a coefficient infront of the bracket
1)2(2 2 xy2xy
1)2( 2 xy
2xy
-1
-2
1
2
3
4
5
-1-2-3-4 1 2 30X->
|̂Y
1)2(2 2 xy
The 2 outside the bracket has stretched the curve vertically by a factor of 2
Translations
4 Sketch the curve found by translating2xy
3
2
2
12xy
by . What is its equation?
5 Sketch the curve found by translating
by . What is its equation?
32 2 )(xy
21 2 )(xy
Translations and Completing the Square
We often multiply out the brackets as follows: 35 2 )(xye.g.
3
355 ))(( xxy
28102 xxy
y x5x5 252x
A quadratic function which is written in the form qpxy 2)(is said to be in its completed square form.
This means multiply ( x – 5 ) by itself
35 2 )(xySo 28102 xx
Completing the Square
The completed square form of a quadratic function
• writes the equation so we can see the translation from
2xy • gives the vertex
Completing the Square
e.g. Consider translated by 2 to the left and 3 up.
2xy
The equation of the curve is 32 2 )(xy
Check: The vertex is ( -2, 3)
3
2
We can write this in vector form as:
translation
Completed square form
Completing the Square
= 2(x2 + x + x + 1) + 3= 2(x2 + x + x
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
= 2x2 + 4x + 2 + 3
= 2x2 + 4x + 5
= 2(x2 + x
Completing the Square
If the graph is plotted we find that the vertex is at (–1, 3)The graph has a Horizontal translation of –1 so q = +1Opposite sign
Vertical translation of 3 so r = 3 Same Sign
-1
-2
-3
1
2
3
4
5
6
7
8
9
-1-2-3-4 1 2 3 40X->
|̂Y
Y=2x̂ 2+4x+5
2x2 + 4x + 5 = p(x + q)2 + r
p = coefficient of x2 p = 2 So 2x2 + 4x + 5 = p(x + q)2 + r = 2(x + 1)2 + 3
= 2(x + 1)2 + 3
We need to find the values of p, q and r
Completing the Square
Using a Calculator to Complete the Square1)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 SquareEx1 y = 2x2 – 3x – 5
Express in the form p(x + q)2 + r
Using 2ndF Trace button (Calc) to find the vertexMin at x = 0.75 and y = –6.125
-1
-2
-3
-4
-5
-6
-7
-8
-9
1
2
3
4
5
-1-2 1 2 3 40X->
|̂Y
Y=2x̂ 2-3x-5
Horizontal translation of + 0.75 so q = –0.75Opposite sign
Vertical translation of –6.125 so r = –6.125Same Sign
p = coefficient of x2 = 2
So 2x2 – 3x – 5 = 2(x – 0.75)2 – 6.125
Completing the Square
Ex1 y = 4 – 3x – x2
Express in the form p(x + q)2 + r
Using 2ndF Trace button (Calc) to find the vertexMax at x = –1.5 and y = 6.25
Horizontal translation of –1.5 so q = 1.5Opposite sign
Vertical translation of 6.25 so r = 6.25Same Sign
p = coefficient of x2 = –1
So 4 – 3x – x2 = –1(x + 1.5)2 + 6.25
-1
-2
-3
1
2
3
4
5
6
7
8
9
-1-2-3-4 1 2 3 40X->
|̂Y
Y=4-3x-x̂ 2
Completing the Square
642 xx
342 xx
1.
2.
3. 1062 xx
22 2 )(x
72 2 )(x
13 2 )(x
ExercisesComplete the square for the following quadratics: