1. Aim: How do transformations affect the equations and graphs of functions?
Do Now: Graph y = -.25x2 – 4 and describe some of the important features.
Axis of symmetry x = 0
x = 0
Maximum Turning Point (0,-4)
Opens down

2. The absolute value of a determines “fatness”.
Dilation of Parabola
a > 1
In other words the absolute value of a, the coefficient of x2, is a dilation factor that changes the shape of the parabola
a < 1
a = 1
f(x) = ax2 + bx + c
The absolute value of a determines “fatness”.
As the absolute value of a decreases in value, the shape of the parabola gets fatter or wider.
As the absolute value of a increases, the shape of the parabola gets “thinner”.
dilation Da

3. +c -c The Nature of the Parabola A(x, y)  A’(x, y + c)
Translation of Parabola: the “c” effect
f(x) = ax2 + bx + c
y = x2 + 6
(0,6) +c
In a quadratic equation “c” tells where the graph crosses the y-axis.
y = x2
(0,0)
The Nature of the
Parabola
y = x2 – 3
(0,-3)
y = x2 – 5
(0,-5) -c
Translation T0,c
A(x, y)  A’(x, y + c)
f(x)  f(x) + c

4. Reflection of a parabola
On the same set of axes, graph y = .25x2 – 4 and describe your results:
Reflection of a parabola
Turning point - (0, -4)
Axis of symmetry x = 0
y = .25x2 – 4
Observations:
(-6,5)
(-6,-13)
Same turning point & axis of symmetry
9 units
(-4,0)
(-4,-8)
(4,0)
(4,-8)
other symmetry?
4 units
(-2,-3)
(-2,-5)
y = -4
y = .25x2 – 4 is a reflection of y = -.25x2 – 4 about the horizontal line whose equation is y = -4
y = -.25x2 – 4
Each point on one graph is equidistant from the line of reflection (y = -4) as its image on the other reflected graph.

5. Graph y = -x2 + 6x – 1 and reflect it about the y-axis
Axis of Symmetry
x = -6/2(-1) = 3
Table of Values
(3,8)
(2,7)
(4,7)
-(0)2 + 6(0) ,-1
(1,4)
(5,4)
-(1)2 + 6(1) ,4
-(2)2 + 6(2) ,7
(0,-1)
(6,-1)
-(3)2 + 6(3) ,8
-(4)2 + 6(4) ,7
-(5)2 + 6(5) ,4
-(6)2 + 6(6) ,-1
Algebra of Reflection: image of A(x,y) in the y-axis is A’(-x, y) image of B(x,y) in the x-axis is B’(x, -y) To graph: find the coordinates of the image of each of the seven points under a reflection in the y-axis and connect those points with a smooth line.

6. Graph y = -x2 + 6x – 1 and reflect it about the y-axis
Image of A(x, y) in the y-axis is A’(-x, y)
f(x)  f(-x)
Reflection
x = - 3
(5,4)
(6,-1)
(2,7)
(1,4)
(0,-1)
x = 3
(4,7)
(3,8)
(-3,8)
(x, y)  (-x, y)
(-4,7)
(-2,7)
(0,-1)
(0,-1)
(-5,4)
(-1,4)
(1,4)
(-1,4)
(2,7)
(-2,7)
(3,8)
(-3,8)
(-6,-1)
(0,-1)
(4,7)
(-4,7)
(5,4)
(-5,4)
(6,-1)
(-6,-1)
Note: axis of symmetry is also reflected under the same rules
Algebra of Reflection: image of A(x,y) in the y-axis is A’(-x, y) image of B(x,y) in the x-axis is B’(x, -y) To graph: find the coordinates of the image of key points under a reflection in the y-axis and connect those points with a smooth line.
What is the equation of the reflected parabola?
y = -x2 – 6x – 1

7. 1. 2. 3. 4.
Which of the above graphs represent f(x) = x2 + 1? 1
Which represents the image of the parabola f(x) = x2 + 1 under a reflection in the:
A. x-axis
B. y-axis
C. The line y = x
D. The line y = -x 3 1 2 4

8. Graph the translation of y = -x2 + 6x – 1 under a translation that maps (x, y)  (x – 2, y + 1)
The image of every point on y = -x2 + 6x – 1 is two units to the left and one unit up.
(x, y)  (x – 2, y + 1)
(-1,5) 
(-2,0)
(1,9)
(0,-1)
(1,4)
(2,7)
(3,8)
(4,7)
(5,4)
(6,-1)
(-2, 0)
(-1, 5)
(0, 8)
(1, 9)
(2, 8)
(3, 5)
(4, 0)
(3,8)
(4,7)
(5,4)
(1,4)
(0,-1)
(6,-1)
Algebra of Translation: image of A(x, y)  A’(x + a, y + b); a is the change in horizontal unit distance and b is the change in vertical unit distance To graph: find the coordinates of the image and connect those points with a smooth line.

9. Vertex Form of Equations of Parabola
Graph the following parabola
on the same set of axes:
y = x2
y = x2 – 5
y = (x + 3)2 - 5
y = (x – 5)2 - 5
y = (x + 7)2 + 1
Describe your findings:

10. Vertex Form of Equations of Parabola
y = (x + 7)2 +1
Vertex Form of Equations of Parabola
f(x) = x2
y = (x + 3)2 - 5
y = (x - 5)2 – 5
f(x) = x2 – 5
Trans. Rule?
A(x, y)  A’(x , y – 5)
f(x) = x2
f(x) = (x + 3)2 – 5
A(x, y)  A’(x – 3, y – 5)
f(x) = (x – 5)2 – 5
A(x, y)  A’(x + 5, y – 5)
y = x2 – 5
f(x) = (x + 7)2 + 1
A(x, y)  A’(x – 7, y + 1)
f(x) = (x – h)2 + k represents a horizontal
translation of f(x) = x2, h units to the right if h is positive or h units to left if h is negative and a vertical translation of k units. The coordinate (h, k) is the turning point of the parabola.

11. Aim: How do transformations affect the equations and graphs of functions?
Do Now:
Write the equation and sketch the graph of y = x3 after a transformation that translates it 3 units up.
…… after a transformation that translates it 3 to the left.
…… after both: a transformation that translates it 3 units up and 3 to the left.

12. Vertical & Horizontal Translations
If k and h are positive numbers and
f(x) is a function, then
f(x) + k shifts f(x) up k units
f(x) – k shifts f(x) down k units
f(x – h) shifts f(x) right h units
f(x + h) shifts f(x) left h units
f(x) = (x – h)2 + k - quadratic
f(x) = |x – h| + k absolute value
f(x) = (x – h)3 + k - cubic
f(x) = (x – h)4 + k - quartic or 4th degree
ex. f(x) = (x – 4)2 + 4 is the image of g(x) = x2
after a shift of 4 units to the right and four
units up or a translation of T4,4.

13. Reflections of Functions
image of B(x,y) in the x-axis is B’(x, -y)
Given f(x):
-f(x) is a reflection of f(x) through the x-axis
f(x)
= (x – 3)2 + 1
f(x) = (x – h)2 + k - parabolic
(3, 1)
(h, k) is the turning point of the parabola.
-f(x)
= -[(x – 3)2 + 1]

14. Reflections of Functions
image of A(x,y) in the y-axis is A’(-x, y)
Given f(x):
f(-x) is a reflection of f(x) through the y-axis
f(-x)
f(x)
= ((-x) – 3)2 + 1
= (x – 3)2 + 1

15. Reflections of Functions
Under reflection in the origin, the Image of P(x, y)  P”(-x, -y)
Given f(x):
-f(-x) is a reflection of f(x) through the origin
= -((-x) – 3)2 + 1)
f(x)
-f(-x)
= (x – 3)2 + 1

16. Dilations of Functions
Given f(x):
af(x) is a dilation of f(x) by a factor of a
If a > 1, the function is ‘stretched’ vertically
2(f(x)) = 2[(x – 3)2 + 1]
.5(f(x)) = .5[(x – 3)2 + 1]
If a < 1, the function is ‘stretched’ horizontally
f(x)
= (x – 3)2 + 1

17. Transformation of Functions
Translation
Reflection
Dilation
f(x) = (x ± h)n ± k n is integer > 1
-f(x) is a reflection of f(x) through the x-axis
f(-x) is a reflection of f(x) through the y-axis
-f(-x) is a reflection of f(x) through the origin
af(x) is a dilation of f(x) by a factor of a
If a > 1, the function is ‘stretched’ vertically
If a < 1, the function is ‘stretched’ horizontally

18. Regents Prep

19. Model Problems
Graph: y = (x – 1)2 – 5

20. Standard to Vertex Form
Rewrite the equation of a parabola in vertex form. f(x) = x2 + 4x + 1
f(x) = x2 + 4x + 1
Separate the first two terms from c.
f(x) = (x2 + 4x ) + 1
Take 1/2 the coefficient of the b term, square it then add the result to the terms inside the parentheses and subtract it from the constant outside.
f(x) = (x2 + 4x ) + 1
– 4
Rewrite the perfect square trinomial in the parenthesis as a binomial square and add the constants together.
f(x) = (x + 2)2 – 3

21. Translating Absolute Value Functions
Graph y = | x |
y = | x – 3 | y = | x + 3 |
y = | x – 3 |
y = | x + 3 |
y = | x |
y = |x + h| represents a horizontal
translation of y = |x|, h units to the left if h is positive or h units to right if h is negative

22. Translating Absolute Value Functions
Graph y = | x |
Graph y = | x | + 3 and y = | x | – 2
y = | x | + 3
(0,3)
y = | x |
(0,0)
y = | x | – 2
(0,-2)
y = |x| + k represents a vertical
translation of y = |x|, k units

23. Translating Absolute Value Functions
Graph y = | x – 3 | + 1
y = | x – 3| + 1
y = | x |
(0,0)
y = | x – 3 |
y = |x – h| + k represents a horizontal shift of h units to the right if h is positive or h units to left if h is negative and a vertical
translation of k units

24. Translating Absolute Value Functions
standard form -
Graph y = | x – 3 |
a = 1, b = 1, c = -3, and d = 0
Find the coordinate of the vertex by evaluating bx + c = 0
bx + c = 0  1(x) – 3 = 0
x = 3
Construct a table of values
using the x-value of the
vertex and several values
to the left and right of it. x
|x – 3| y
x,y 1 2 3 4 5 ,3 ,2 ,1
Graph the resulting points
to form the V-shaped graph ,0 ,1
If a is positive V opens up ,2
If a is negative V opens down

25. The Punted Football
The height of a punted football can be
modeled by the quadratic function
h = x x The horizontal distance
in feet from the point of impact is x, and h is the
height of the ball in feet.
a. Find the vertex of the graph of the function
by completing the square.
What is the maximum height of this punt?
c. The nearest defensive player from the point
of impact is 5 feet away. How high must he get
his hand to block the punt?

26. The Punted Football Really Small Peoples’ Football League x = 5
h = x x + 2
x = 5 4’ 2’

27. The Punted Football (x - 59)2 = 3681 x - 59 = ±60.67 x – 59 = 60.67
h = -x x + 200
Rewrite the quadratic
for h = 0:
x2 – 118x = 200
Take 1/2 the coefficient
of the linear term &
square it. This is now the c term.
= 3481
Add the c term to
both sides of
equation
x2 – 118x = 200
+ 3481
+ 3481
(x - 59)2 = 3681
Binomial Squared
Find square root
of both sides
Solve for x
x - 59 = ±60.67
x – 59 = 60.67
x – 59 =
x =
x = -1.67