Download - UNIT 1 REVIEW of TRANSFORMATIONS of a GRAPH
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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
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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
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FIND THE EQUATION for graphs #5, 8:
Graph #5:y = a(x – h)2 + k
Graph #8:y = a(x – h)2 + k
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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
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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
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FIND THE EQUATION for Graphs #7 ,9:Graph #7: Graph #9:
khxay khxay
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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?
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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)
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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
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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
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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)
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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
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[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
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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
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[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
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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
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[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
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[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
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[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
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[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