Download Warm-Up (Over Chapter 6)

UNIT 1 REVIEW of TRANSFORMATIONS of a GRAPH
f(x)
a•f(x – h) + k
Original (Parent) Graph
Transformed Graph
“a” – value
“h” – value
“k” – value
For graphs #1, 2, 3, 5, 8:
y = x2
y = a(x – h)2 + k
QUADRATIC FUNCTION
(ORIGINAL)
TRANSFORMED
QUADRATIC FUNCTION
FIND THE EQUATION for Graphs #1, 2, 3:
Step #1: Use the vertex to indicate the horizontal (h) and
vertical (k) changes
Step #2: Use another point on the graph to help
determine the “a”- value (Hint: The direction it opens indicates the sign)
Graph #3:
Graph #2:
Graph #1:
2+k
2
2
y
=
a(x
–
h)
y
=
a(x
–
h)
+
k
y = a(x – h) + 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”
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?
y
ya xhk
x
TRANSFORMED
SQUARE ROOT FUNCTION
SQUARE ROOT FUNCTION
(ORIGINAL)
“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.
Step #2: Use another point on the graph to determine the “a”- value
Step #3: Solve the transformed equation for x
Graph #4:
ya xhk
Graph #6:
ya xhk
FIND THE EQUATION for Graphs #7 ,9:
Graph #7:
Graph #9:
ya xhk
ya xhk
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 Findings
Parabolas
Parabolas
(Type 2: Right and Left)
(Type 1: Up and Down)
Vertex Form
Vertex Form
y  a( x  h)  k
2
Vertex: (h, k)
x  a( y  k )  h
2
Vertex: (h, k)
Axis: x = h
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 Vertical
[B] Opening: Horizontal
Vertex: (2, 4)
Vertex: (- 4, 6)
Point: (-6, 8)
Point: (2, 8)
y  a( x  h)2  k
y  a ( x  2) 2  4
x  a( y  k ) 2  h
x  a( y  6)2  4
8  a ( 6  2) 2  4
8  64a  4
1
a
16
1
y  ( x  2)2  4
16
2  a(8  6)2  4
6  4a
3
a
2
3
x  ( y  6)2  4
2
COMPLETING THE SQUARE REVIEW
Find the value to add to the trinomial to create a
perfect square trinomial: (Half of “b”)2
[A]
x  10 x  c
2
[B]
x  5x  c
2
x  10 x  ___
x 2  5 x  ___
(10  2)  25
( 5  2) 2  25 4
x  10 x  25
x 2  5 x  25 4
( x  5) 2
( x  5 2 )2
2
2
2
[C] 2 x  8 x  c
2( x 2  4 x  ___)
2
[D] 3 x  9 x  c
3( x 2  3 x  ___)
2
( 4  2)  4
( 3  2) 2  9 4
2( x  4 x  4)
3( x 2  3 x  9 4 )
2
2
2( x  2)  2 x  8 x  8
2
2
3( x  3 2 )2  3 x 2  9 x  27 4
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
a 1
Write in vertex form. Identify the vertex and axis of symmetry.
y  x  6x  8
2
[A]
6
 3;3 2  9
2
y  ( x 2  6 x  9)  3  9
y  ( x  3) 2  6
Vertex: (-3, -6)
Axis:
x = -3
[B]
y  x  4x  3
2
4
 2; ( 2)2  4
2
y  ( x 2  4 x  4)  3  4
y  ( x  2) 2  1
Vertex:
(2, -1)
Axis:
x=2
Example 2 Type 2: Right or Left Parabolas
a 1
Write in vertex form. Identify the vertex and axis of symmetry.
[A]
x  y  8y  8
2
8
 4; (4)2  16
2
x  ( y 2  8 y  16)  8  16
x  ( y  4)  8
2
Vertex:
(-8, -4)
Axis:
y = -4
[B]
x  y2  6 y  4
6
 3; ( 3)2  9
2
x  ( y 2  6 y  9)  4  9
x  ( y  3) 2  5
Vertex:
(-5, 3)
Axis:
y=3
a 1
Example 3 Type 1: Up or Down Parabolas
Write in standard form. Identify the vertex and axis of symmetry.
[A]
y  3 x  24 x  50
2
y  3( x 2  8 x )  50
8
 4; (4) 2  16
2
y  3( x 2  8 x  16)  50  ( 3)  16
y  3( x  4) 2  2
Vertex:
(-1, 4)
Axis:
x = -1
[B] y   x  2 x  3
2
 1; (1)2  1
2
y  ( x 2  2 x  1)  3  ( 1)  1
2
y  ( x  1)2  4
Vertex:
(-1, 4)
Axis:
x = -1
Example 4 Type 2: Right or Left Parabolas
a 1
Write in vertex form. Identify the vertex and axis of symmetry.
[A]
x  5 y 2  25 y  7
x  5( y  5 y )  7
5 5 5 2 25
 ;( ) 
2 2 2
4
25
25
2
x  5( y  5 y  )  7  (5)
4
4
25
97
x  5( y  ) 2 
2
4
2
Vertex:
Axis:
(-25/2, -97/4 )
y = -97/4
[B]
x  3 y  12 y  1
2
x  3( y 2  4 y )  1
4
 2; ( 2) 2  4
2
x  3( y 2  4 y  4)  1  ( 3)4
x  3( y  2) 2  13
Vertex:
Axis:
(13, -2)
y = -2
Method #2: SHORTCUT
1. Find the AXIS of SYMMETRY :
Axis is horizontal or vertical based on shape
b
x
2a
b
y
2a
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)
PRACTICE METHOD #2: Slide 2
[1]
Write in vertex form. Find vertex and axis of symmetry.
2
2
y  x  8x  1
[2] y  x  10 x  20
 b  ( 8)

 4  x
2a
2(1)
y  ( 4) 2  8( 4)  1  15
a 1
y  ( x  4) 2  15
Vertex: (- 4, -15)
Axis: x = - 4
 b  (10)

 5  x
2a
2( 2)
y  ( 5) 2  10( 5)  20  5
a 1
y  ( x  5) 2  5
Vertex: (5, - 5)
Axis: x = 5
PRACTICE METHOD #2: Slide 2
Write in vertex form. Find vertex and axis of symmetry.
2
[3] y  3 x 2  6 x  5
[4] y  2 x  16 x  32
 b  ( 6)

1 x
2a
2( 3)
y  3(1) 2  6(1)  5  2
a3
y  3( x  1) 2  2
Vertex: (1, 2)
Axis: x = 1
 b  ( 16)

 4  x
2a
2( 2)
y  2( 4) 2  16( 4)  32  0
a  2
y  2( x  4) 2
Vertex: (-4, 0)
Axis: x = -4
PRACTICE METHOD #2: Slide 3
Write in vertex form. Find vertex and axis of symmetry.
2
[5] x  y 2  4 y  7
[6] x  y  5 y  4
 b  ( 4)

2 y
2a
2(1)
x  ( 2) 2  4( 2)  7  3
a 1
x  ( y  2) 2  3
Vertex: (3, -2)
Axis: y = -2
 b  ( 5)
5

  y
2a 2(1)
2
5 2
5
41
y  (  )  5(  )  4  
2
2
4
a 1
5 2 41
x  (y ) 
2
4
Vertex: (-41/5, - 5/2)
Axis: y = -5/2
PRACTICE METHOD #2: Slide 4
Write in vertex form. Find vertex and axis of symmetry.
2
[7] x  4 y 2  16 y  13
[8] x  3 y  9 y  1
 b  (16)

 2  y
2a
2(4)
x  4( 2) 2  16( 2)  13  3
a4
x  4( y  2) 2  3
Vertex: (-3, -2)
Axis: y = -2
 b  ( 9 )
3

  y
2a 2( 3)
2
3 2
3
31
x  3(  )  9(  )  1 
2
2
4
a  3
3 2 31
x  3( y  ) 
2
4
Vertex: (31/4, -3/2)
Axis: y = -3/2