Aim: Transforming Functions 1 Course: Alg. 2 & Trig. Aim: How do transformations affect the...

Aim: Transforming Functions 1 Course: Alg. 2 & Trig.

Aim: How do transformations affect the equations and graphs of functions?

Do Now: Graph y = -.25x2 – 4 and describe some of the important features.

Maximum Turning Point (0,-4)

Axis of symmetry x = 0

x =

0

Opens down


The absolute value of a determines “fatness”.

f(x) = ax2 + bx + ca = 1

a < 1

a > 1

As the absolute value of a increases, the shape of the parabola gets “thinner”.

As the absolute value of a decreases in value, the shape of the parabola gets fatter or wider.

Dilation of Parabola

In other words the absolute value of a,

the coefficient of x2, is a dilation factor that changes the shape of

the parabola

dilation Da


Translation of Parabola: the “c” effect

f(x) = ax2 + bx + c

Translation of Parabola: the “c” effect

f(x) = ax2 + bx + c

+c

In a quadratic equation

“c” tells where the

graph crosses the

y-axis.

-c

y = x2 + 6(0,6)

y = x2(0,0)

y = x2 – 3(0,-3)

y = x2 – 5(0,-5)

A(x, y) A’(x, y + c) f(x) f(x) + c

Translation T0,c


Reflection of a parabolaOn the same set of axes, graph y = .25x2 – 4 and describe your results:

Same turning point & axis of symmetry

4 units

4 units

(4,0)

(4,-8)

(-2,-3)

(-2,-5)

(-4,0)

(-4,-8)

Each point on one graph is equidistant from the line of reflection (y = -4) as its image on the other reflected graph.

9 units

9 units

Axis of symmetry x = 0

(-6,5)

(-6,-13)

y = .25x2 – 4 is a reflection of

y = -.25x2 – 4 about the

horizontal line whose equation

is y = -4

y = -4

Observations:

Turning point - (0, -4)

y = -.25x2 – 4

y = .25x2 – 4

other symmetry?


(4,7)

(5,4)

(6,-1)

Graph y = -x2 + 6x – 1 and reflect it about the y-axis

Table of Values

-(0)2 + 6(0) - 1 -1 0,-1

-(1)2 + 6(1) - 1 4 1,4

-(2)2 + 6(2) - 1 7 2,7

-(3)2 + 6(3) - 1 8 3,8

-(4)2 + 6(4) - 1 7 4,7

-(5)2 + 6(5) - 1 4 5,4

-(6)2 + 6(6) - 1 -1 6,-1

Axis of Symmetry

x b

2ax = -6/2(-1) = 3

(2,7)

(1,4)

(0,-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.


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.

(3,8)


(x, y) (-x, y)(0,-1)

(1,4)

(2,7)

(3,8)

(4,7)

(5,4)

(6,-1)

(0,-1)

(-1,4)

(-2,7)

(-3,8)

(-4,7)

(-5,4)

(-6,-1)

Reflection

Graph y = -x2 + 6x – 1 and reflect it about the y-axis

(-5,4)

(-2,7)(-4,7)

(-1,4)

(0,-1)(-6,-1)

Image of A(x, y) in the y-axis is A’(-x, y)

Note: axis of symmetry is also reflected under the same rules

x = - 3

What is the equation of the reflected parabola?y = -x2 – 6x – 1

(5,4)

(6,-1)

(2,7)

(1,4)

(0,-1)

x = 3

(4,7)(3,8)(-3,8)

f(x) f(-x)


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.


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.


6

4

2

-2

-4

-6

-5 5

6

4

2

-2

-4

-6

-5 5

6

4

2

-2

-4

-6

-5 5

6

4

2

-2

-4

-6

-5 5

Which of the above graphs represent f(x) = x2 + 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

1. 2. 3. 4.

1

3

1

2

4

6

4

2

-2

-4

-6

-5 5


(x, y) (x – 2, y + 1)

Graph the translation of y = -x2 + 6x – 1 under a translation that maps (x, y) (x – 2, y + 1)

(0,-1)

(1,4)

(2,7)

(3,8)

(4,7)

(5,4)

(6,-1)

(1,4)

(0,-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.

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.

(-2, 0)

(-1, 5)

(0, 8)

(1, 9)

(2, 8)

(3, 5)

(4, 0)

(3,8)

(-1,5)

(-2,0)

(1,9)

The image of every point on y = -x2 + 6x – 1 is two units to the left and one unit up.

(6,-1)

(5,4)

(4,7)


Vertex Form of Equations of ParabolaVertex Form of Equations of Parabola

y = x2 – 5

y = (x + 7)2 + 1

y = (x – 5)2 - 5

y = (x + 3)2 - 5

Graph the following parabola on the same set of axes:

Describe your findings:

y = x2


y =

x2 –

5

y =

(x

+ 7

)2 +

1f(x) = (x + 7)2 + 1

A(x, y) A’(x – 7, y + 1)y

= (

x +

3)2

- 5

f(x) = (x + 3)2 – 5

A(x, y) A’(x – 3, y – 5)

f(x) = x2

f(x) = (x – h)2 + k represents a horizontaltranslation 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.

f(x) = x2

y =

(x

- 5)

2 –

5f(x) = (x – 5)2 – 5

Vertex Form of Equations of ParabolaVertex Form of Equations of Parabola

f(x) = x2 – 5

A(x, y) A’(x , y – 5)

A(x, y) A’(x + 5, y – 5)

Trans. Rule?


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.


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 - quadraticf(x) = |x – h| + k - absolute value

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.

f(x) = (x – h)3 + k - cubic

f(x) = (x – h)4 + k - quartic or 4th degree


6

4

2

-2

-4

-6

5

Reflections of Functions

Given f(x):-f(x) is a reflection of f(x) through the x-axis

f(x)

-f(x)

= (x – 3)2 + 1

= -[(x – 3)2 + 1]

image of B(x,y) in the x-axis is B’(x, -y)

(h, k) is the turning point of the parabola.

(3, 1)

f(x) = (x – h)2 + k - parabolic


6

4

2

-2

-4

-6

-5 5


Given f(x):f(-x) is a reflection of f(x) through the y-axis

f(x)

= (x – 3)2 + 1

f(-x)

= ((-x) – 3)2 + 1

image of A(x,y) in the y-axis is A’(-x, y)


6

4

2

-2

-4

-6

-5 5


Given f(x):-f(-x) is a reflection of f(x) through the origin

f(x)

= (x – 3)2 + 1-f(-x)

= -((-x) – 3)2 + 1)

Under reflection in the origin, the Image of P(x, y) P”(-x, -y)


6

4

2

-2

-4

-6

5

Dilations of FunctionsGiven f(x):

af(x) is a dilation of f(x) by a factor of a

f(x) = (x – 3)2 + 1

2(f(x)) = 2[(x – 3)2 + 1]

.5(f(x)) = .5[(x – 3)2 + 1]

If a < 1, the function is ‘stretched’ horizontally

If a > 1, the function is ‘stretched’ vertically


Translation

Reflection

Dilation

Transformation of Functions

af(x) is a dilation of f(x) by a factor of a

If a < 1, the function is ‘stretched’ horizontallyIf a > 1, the function is ‘stretched’ vertically

-f(-x) is a reflection of f(x) through the origin

f(-x) is a reflection of f(x) through the y-axis

-f(x) is a reflection of f(x) through the x-axis

f(x) = (x ± h)n ± k n is integer > 1


Regents PrepRegents Prep

, sketch the graph of

the transformation of

If ( ) 2

( ) 4. What are

the minimum values of and of the

transformed function?

f x x

f x

x y


Model Problems

Graph: y = (x – 1)2 – 5

8

6

4

2

-2

-4

-6

-8

-10 -5 5 10


Standard to Vertex Form

Rewrite the equation of a parabola in vertex form. 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.

Separate the first two terms from c.

f(x) = (x2 + 4x ) + 1

f(x) = (x2 + 4x ) + 1 + 4 – 4

Rewrite the perfect square trinomial in the parenthesis as a binomial square and add the constants together.

f(x) = (x + 2)2 – 3

8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

f(x) = x2 + 4x + 1


Translating Absolute Value Functions

y = | x – 3 | Graph y = | x | y = | x – 3 | y = | x + 3 |

y = |x + h| represents a horizontaltranslation of y = |x|, h units to the left if h is

positive or h units to right if h is negative

y = | x + 3 | y = | x |



Graph y = | x |

Graph y = | x | + 3 and y = | x | – 2

y = |x| + k represents a vertical translation of y = |x|, k units

y = | x | – 2

(0,-2)

y = | x |

(0,0)

y = | x | + 3

(0,3)



Graph y = | x – 3 | + 1

y = | x – 3| + 1

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

y = | x – 3 | y = | x |

(0,0)



Graph y = | x – 3 |

standard form -

y a bx c d

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 valuesusing the x-value of the vertex and several valuesto the left and right of it.

Graph the resulting pointsto form the V-shaped graph

x |x – 3| y x,y

0

1

2

3

4

5If a is positive V opens up

If a is negative V opens down

3 3 0,3

2 2 1,2

1 1 2,1

0 1 3,0

1 1 4,1

2 2 5,2


The Punted Football

The height of a punted football can bemodeled by the quadratic function h = - 0.01x2 + 1.18x + 2. The horizontal distancein feet from the point of impact is x, and h is theheight of the ball in feet.

a. Find the vertex of the graph of the functionby completing the square.

b. What is the maximum height of this punt?

c. The nearest defensive player from the pointof impact is 5 feet away. How high must he gethis hand to block the punt?


The Punted Football

2’

4’

h = - 0.01x2 + 1.18x + 2

x = 5

Really Small Peoples’Football League


The Punted Football

Solve for x

Find square rootof both sides

Binomial Squared

Add the c term to both sides of equation

Take 1/2 the coefficientof the linear term & square it. This is now the c term.

h = -x2 + 118x + 200

x2 – 118x = 200

( 59)2

+ 3481 + 3481

(x - 59)2 = 3681

(x 59)2 3681

x - 59 = ±60.67

x – 59 = 60.67 x – 59 = -60.67

x = 119.67 x = -1.67

= 3481

Rewrite the quadraticfor h = 0: x2 – 118x = 200

Aim: Transforming Functions 1 Course: Alg. 2 & Trig. Aim: How do transformations affect the...

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