Translations and Completing the Square © Christine Crisp.

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Translations and Translations and Completing the Square Completing the Square © Christine Crisp

Transcript of Translations and Completing the Square © Christine Crisp.

Translations and Translations and Completing the SquareCompleting 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

SUMMARY

The curve

is a translation of by 2xy

q

pqpxy 2)(

The vertex is given by ),( qp

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:

Completing the Square

282 xx

332 xx

182 2 xx

184 2 )(x

432

23 )(x

722 2 )(x

4.

5.

6.

Qu.s in notes pg 32