UNIT 1 REVIEW of TRANSFORMATIONS of a GRAPHf(x)

Original (Parent) Grapha•f(x – h) + kTransformed Graph

“a” – value

“h” – value

“k” – value

For graphs #1, 2, 3, 5, 8:y = x2

QUADRATIC FUNCTION(ORIGINAL)

y = a(x – h)2 + kTRANSFORMED

QUADRATIC FUNCTION

FIND THE EQUATION for Graphs #1, 2, 3:Step #1: Use the vertex to indicate the horizontal (h) and vertical (k) changes

Graph #1:y = a(x – h)2 + k

Step #2: Use another point on the graph to help determine the “a”- value (Hint: The direction it opens indicates the sign)

Graph #2:y = a(x – h)2 + k

Graph #3:y = a(x – h)2 + k

FIND THE EQUATION for graphs #5, 8:

Graph #5:y = a(x – h)2 + k

Graph #8:y = a(x – h)2 + k

For graphs #4, 6, 7, and 9:“Sideways Quadratic Graphs”

SQUARE ROOT FUNCTION(ORIGINAL)

TRANSFORMED SQUARE ROOT FUNCTION

1) If you look at the symmetrical parts of these graphs above or below the axis of symmetry, what function do these parts most resemble?

2) Write down the general equation of the parent function and transformed function?

xy khxay

“a” – value “h” – value “k” – value

FIND THE EQUATION for Graphs #4 ,6:Step #1: Use the vertex of the graph to indicate the horizontal (h) and vertical (k) changes for the starting pt.

Graph #4:

Step #2: Use another point on the graph to determine the “a”- value

Graph #6:

khxay

Step #3: Solve the transformed equation for x

khxay

FIND THE EQUATION for Graphs #7 ,9:Graph #7: Graph #9:

khxay khxay

OBSERVATIONS: Look at the equations for graphs #4, 6, 7, 9

1)Do you notice any SIMILARITIES or DIFFERENCES in those equations in comparison to the quadratic?

2)How are the coordinates of the VERTEX in the graph related to the equation?

3)How is the axis of symmetry equation and vertex related based on the shape of the graph?

Parabola Formulas Summary of Day One FindingsParabolas

(Type 2: Right and Left)Parabolas

(Type 1: Up and Down)

Vertex Form Vertex Form

khxay 2)( hkyax 2)(

Vertex: (h, k)

Axis: x = h

Vertex: (h, k)

Axis: y = k

Rate: a (+ up; – down) Rate: a (+ right; –left)

Find VERTEX FORM EQUATION: Given Vertex & Point

Plug vertex into appropriate vertex form equation and use another point to solve for “a”.

[A] Opening VerticalVertex: (2, 4)Point: (-6, 8)

[B] Opening: HorizontalVertex: (- 4, 6)Point: (2, 8)

khxay 2)( hkyax 2)(

4)6( 2 yax

4)68(2 2 a

2

3

46

a

a

4)6(2

3 2 yx

a

a

a

xay

16

1

4648

4)26(8

4)2(2

2

4)2(16

1 2 xy

COMPLETING THE SQUARE REVIEWFind the value to add to the trinomial to create a

perfect square trinomial: (Half of “b”)2

[A] cxx 102[B] cxx 52

[C] cxx 82 2[D] cxx 93 2

882)2(2

)44(2

4)24(

___)4(2

22

2

2

2

xxx

xx

xx

2

2

2

2

)5(

2510

25)210(

___10

x

xx

xx

22

5

4252

4252

2

)(

5

)25(

___5

x

xx

xx

42722

23

492

492

2

93)(3

)3(3

)23(

___)3(3

xxx

xx

xx

VERTEX FORM: DAY TWO

FIND VERTEX FORM given STANDARD FORM

Method #1: COMPLETING THE SQUARE•Find the value to make a perfect square trinomial to the quadratic equation.

(Be careful of coefficient for x2 which needs to be distributed out)

•ADD ZERO by adding and subtracting the value to make a perfect square trinomial so as to not change the overall equation(Be careful of coefficient for x2 needs multiply by subtraction)

Example 1 Type 1: Up or Down Parabolas Write in vertex form. Identify the vertex and axis of symmetry.

[A] 862 xxy [B] 342 xxy

1a

Vertex: (2, -1)

Axis: x = 2

1)2(

43)44(

4)2(;22

4

2

2

2

xy

xxy

6)3(

93)96(

93;32

6

2

2

2

xy

xxy

Vertex: (-3, -6)

Axis: x = -3

[A] 882 yyx [B] 462 yyx

Vertex: (-5, 3)

Axis: y = 3

5)3(

94)96(

9)3(;32

6

2

2

2

yx

yyx

Example 2 Type 2: Right or Left Parabolas Write in vertex form. Identify the vertex and axis of symmetry.

1a

Vertex: (-8, -4)

Axis: y = -4

8)4(

168)168(

16)4(;42

8

2

2

2

yx

yyx

Write in standard form. Identify the vertex and axis of symmetry.

[A] 50243 2 xxy [B] 322 xxy

1a

Vertex: (-1, 4)

Axis: x = -1

4)1(

1)1(3)12(

1)1(;12

2

2

2

2

xy

xxy

Example 3 Type 1: Up or Down Parabolas

Vertex: (-1, 4)

Axis: x = -1

2)4(3

16)3(50)168(3

16)4(;42

8

50)8(3

2

2

2

2

xy

xxy

xxy

[A] 7255 2 yyx [B] 1123 2 yyx

1a

Vertex: (13, -2)

Axis: y = -2

13)2(3

4)3(1)44(3

4)2(;22

4

1)4(3

2

2

2

2

yx

yyx

yyx

Example 4 Type 2: Right or Left Parabolas Write in vertex form. Identify the vertex and axis of symmetry.

Vertex: (-25/2, -97/4 )

Axis: y = -97/4

4

97)

2

25(5

4

25)5(7)

4

255(5

4

25)

2

5(;

2

5

2

5

7)5(5

2

2

2

2

yx

yyx

yyx

Method #2: SHORTCUT1.Find the AXIS of SYMMETRY :Axis is horizontal or vertical based on shape

2.Find VERTEX (h, k) of STANDARD FORM

3.“a” – value for vertex form should be the same coefficient of x2 in standard form. Check by using another point (intercept)

a

bx

2

a

by

2

[1] 182 xxy

PRACTICE METHOD #2: Slide 2Write in vertex form. Find vertex and axis of symmetry.

Vertex: (- 4, -15)

Axis: x = - 4

15)4(

1

151)4(8)4(

4)1(2

)8(

2

2

2

xy

a

y

xa

b

[2] 20102 xxy

Vertex: (5, - 5)

Axis: x = 5

5)5(

1

520)5(10)5(

5)2(2

)10(

2

2

2

xy

a

y

xa

b

[3] 563 2 xxy

PRACTICE METHOD #2: Slide 2Write in vertex form. Find vertex and axis of symmetry.

Vertex: (1, 2)

Axis: x = 1

2)1(3

3

25)1(6)1(3

1)3(2

)6(

2

2

2

xy

a

y

xa

b

[4] 32162 2 xxy

Vertex: (-4, 0)

Axis: x = -4

2

2

)4(2

2

032)4(16)4(2

4)2(2

)16(

2

xy

a

y

xa

b

[5] 742 yyx

PRACTICE METHOD #2: Slide 3Write in vertex form. Find vertex and axis of symmetry.

Vertex: (3, -2)

Axis: y = -2

3)2(

1

37)2(4)2(

2)1(2

)4(

2

2

2

yx

a

x

ya

b

[6] 452 yyx

Vertex: (-41/5, - 5/2)

Axis: y = -5/2

4

41)

2

5(

14

414)

2

5(5)

2

5(

2

5

)1(2

)5(

2

2

2

yx

a

y

ya

b

[7] 13164 2 yyx

PRACTICE METHOD #2: Slide 4Write in vertex form. Find vertex and axis of symmetry.

Vertex: (-3, -2)

Axis: y = -2

3)2(4

4

313)2(16)2(4

2)4(2

)16(

2

2

2

yx

a

x

ya

b

[8] 193 2 yyx

Vertex: (31/4, -3/2)

Axis: y = -3/2

4

31)

2

3(3

34

311)

2

3(9)

2

3(3

2

3

)3(2

)9(

2

2

2

yx

a

x

ya

b