1

*

*Perfect Square Trinomial

*Quadratics that you get by squaring a binomial

*(x-3)2 = (x-3)(x-3) = x2-6x+9

*Factoring A Perfect Square Trinomial results

in two of the same binomials.

*x2+14x+49 = (x+7)2

*

*Multiply the following:

( )( )6 6x x+ +

( )2

1. 6x+ = ( )2

2. 8x − =

212 36x x+ +

( )( )8 8x x− −

216 64x x− +

*

*Factor completely

23. 6 9x x+ + =24. 10 25x x− + =

2 ( 3)x+ 2( 5)x−

*

*Basically, when you look at a perfect-square trinomial there is a relationship between the

coefficient of the x-term (b) and the last term

(c).

*What is the relationship?

*The process of finding the last term of a perfect-square trinomial

is called Completing the Square.

X2 6x 92

)3( += x

2

X2 4x 4

4

2)2( += x

*

1. Identify b from ax2 + bx + c

2. Find value of

3. Square the value of which means find

4. is the value that completes the square.

2

b 2

2

b

2

b

2

2

b

**Find the missing value to complete the square, then factor.

5.

6.

7.

2( 10)x −

23( )

2x +

2 10 ______x x+ +

2 20 ______x x− +

2 3 ______x x+ +

25

1009

4

2( 5)x +

* Steps: Completing the Square.

1.First move “c” to the other side of the equation.

2. If , factor out the “a” value.

3. Rewrite the problem to look like:

4. Complete the square by using: + + = − +

2

2 2

2 2a x b

bax c

b

1a ≠

( )+ + = − +2 ___ ___a x bx c

* Steps Cont…

5. Factor the left side of the equation to a perfect square and simplify the right side.

6. Divide by “a”, then take the square root of both sides to eliminate the square on the left side of the equation. Don’t forget:

7. Solve for x. (Remember there will usually be two different answers)

±

*

x2 + 10x + 8 = 0

(x2 + 10x ) = -8

(x2 + 10x + (5)2) = -8 + 25

(x + 5)2 = 17

(5)2 = 25

175 ±=+x

175 ±−=x

( ) 5102

1 =

3

* 3x2 + 24x + 12 = 0

(x2 + 8x ) = -4

(x2 + 8x + (4)2) = -4 + 16

(x + 4)2 = 12

(4)2 = 16

124 ±=+x

124 ±−=x

( ) 482

1 =

x2 + 8x + 4 = 0

324 ±−=x

3(x2 +8x +4) = 0

3(x2 +8x +4) = 0

3 = 3

*

*You will need one piece of patty paper

*Shade one of the 4 edges of the paper with your pencil

*Place a dot 1-2 inches above the line on your patty paper

*Fold the shaded line up to touch the dot and crease the fold

*Repeat the folding process 50-60 more times.

*Parabola

*A parabola is the set of all points in a A parabola is the set of all points in a A parabola is the set of all points in a A parabola is the set of all points in a plane plane plane plane

that are the same distance from a given point that are the same distance from a given point that are the same distance from a given point that are the same distance from a given point

called a called a called a called a focusfocusfocusfocus (the dot) (the dot) (the dot) (the dot) and a given line called and a given line called and a given line called and a given line called

a a a a directrixdirectrixdirectrixdirectrix (the shaded edge)(the shaded edge)(the shaded edge)(the shaded edge)

*A Parabola is obtained by slicing a double cone A Parabola is obtained by slicing a double cone A Parabola is obtained by slicing a double cone A Parabola is obtained by slicing a double cone

on a slant on a slant on a slant on a slant

Directrix – Shaded side

Focus

Distance of p

Distance of pVertex

Ax

is o

f S

ym

me

try

Focal Chord orLatus Rectum

( ) ( )2

4 p y k x h− = −

• General form

• Standard form

Forms of a Parabola

20Ax Cx Dy E+ + + =

20By Cx Dy E+ + + =

Vertical Horizontal

( ) ( )2

4 p x h y k− = −

Vertical Horizontal

4

Directrix

Ax

is o

f S

ym

me

try

Focus

Vertex(h,k)

Focal Chord orLatus Rectum

4p(y – k) = (x - h)2

Vertex

Dir

ec

trix

Axis of Symmetry

Focal Chord orLatus Rectum

4p(x – h) = (y - k)2

What do you think?*How is the equation for a vertical parabola similar

to a horizontal parabola?

*How are they different?

*How does each parameter(a,h,k)of the equation

change the graph for a:

*Vertical Parabola

*Horizontal Parabola

Graph 3x-y2=8y+31

1. Determine which direction the parabola will open

This one will open left/right since we have y2

2. Use completing the square to get it into the correct form we need for graphing

23 8 31x y y− = +

23 131 8 3yx y= + +−

23 31 ( 8 ___)x y y− = + +

( )2

284 16

2

= =

( )2

416+

23 15 ( 4)x y− = +

23( 5) ( 4)x y+ = +

23 15 ( 4)x y− = +

2 15( 4

1

3)

3x y= + +

4p(x – h) = (y - k)2

1 1 4 3 so

4 3p

p= =

23 ( 4) 15x y= + +

What is the vertex?

(h,k) can be found directly in the model.

How can we find a?

1

4or a

p=

Rewrite the equation as x =

(-5, -4)

*How do I Look?

*What will I look like graphedWhat will I look like graphedWhat will I look like graphedWhat will I look like graphed????

1.

2.

( )21

1 65

y x= − + +

( )2

3 1 2x y= − + +

Will it open up/down or left/right?

Where is the vertex?

Down since it is y = and a is negative(-1,6)

Will it open up/down or left/right?

Where is the vertex?

Left since it is x = and a is negative(2,-1)

Vertex

Focus

AOS

Directrix

Opening - up or down

*

25 3( 1)y x+ = +

22 3( 2 )y x x+ = +

2 223(1) 32 ( 2 )1y x x+ + = + +

( ) ( )21

5 13

y x+ = +

2632

−+= xxy

xxy 6322

+=+

( ),h k p+

y k p= −

x = h

(h, k) (-1, -5)

x = -1

591,

12

− −

61

12y

−=

Reminder: 4p = so p = ?

1

3p = 112

Why?Because p > 0

What is p? Distance between the vertex and the focus

Remember to add or subtract the value

of p to the correct coordinate

5

Vertex

Focus

AOS

Directrix

Opening – left or right

Ex 4:

Put in standard form

26 3x y y= − +

2 3 6x y y− = −

2 29 ( 33 6 )x y y− + = − −+

( )2

6 3x y+ = −

( )2

6 3x y+ = −

( ),h p k+x h p= −

(h, k)

y = k

( )6,3−

3y =

23,3

4

−

25

4x

−=

Why?Because p > 0

Remember to add or subtract the value

of p to the correct coordinate

**Put the equation into standard form

( )2

1 13

2 2y x

+ = − +

24 4 7 2 0y y x+ + + =

24( ____) 2 7 4(____)y y x+ + = − − +

21

2

1

4

21

4 2 62

y x

+ = − −

( )2

21

42

3y x

+ =

− +

*

Find the vertex.

(x, y) � (2, -3) � (h, k)

h = 2

k = -3

4p =

Find 4p (Use the vertex and another point.)

(2, -3) & (1, -1)

(h, k) & (x, y)

( ) ( )2

4 1 3 1 2p − − − = −

( )4 2 (1)

14

2

p

p

=

=

( ) ( )21

3 22

y x+ = −

12

( ) ( )2

4 p y k x h− = −

Ex 7: Write the equation from the graph

Find the vertex.

(x, y) � (-2, -3) � (h, k)

Find 4p. (Use the vertex and another point.)

(-2, -3) & (-1, -2)

(h, k) & (x, y) ( )4 1 1

4 1

p

p

=

=

( ) ( )2

2 3x y+ = + ( ) ( )2

4 1 2 2 3p − − − = − − −

h = -2

k = -3

4p = 1

( ) ( )2

4 p x h y k− = −

* Ex 8: Write the equation given the directrix, y = 4 and the focus (4,-3) - Method 1

( ),h k p+

14

2p

− =

( ),h k p+

y k p= −

Decide which standard form you need to use.

Because the directrix is y= it is the 4p(y-k)=(x-h)2

Find a, h, and k. h is 4 because the focus formula is

To find k and a we need to use both

formulas 3k p+ = −

4k p− =

12 =k

2

1=k

7

2p− =

7

2p = −

4 14p = −

4 14

4

1

2

p

h

k

= −

=

=

( )21

14 42

y x

− − = −

y = 4

(4, -3)

* Ex 8: Write the equation given the directrix, y = 4 and the focus (4,-3) - Method 2

Decide which standard form you need to use.

Because the directrix is y= it is the 4p(y-k)=(x-h)2

Sketch a graph.

( )21

14 42

y x

− − = −

Where is the vertex?

What is p?

at (4, ½) Halfway between the focus and directrix on the AOS.

p = -3.5 or 7/2so, 4p = -14

6

*Ex 9: Communication

*A microphone is placed at the focus of a

parabolic reflector to collect sound for the

television broadcast of the World Cup soccer

game. Write an equation for the cross section,

assuming that the focus is at the origin and the

parabola opens to the right.

6 units

*

* Groups of 3 or 4

* Each group will solve and graph one parabola and write the equation for another. Graphs will be place on large graph paper with work out to the side on white paper.

* Supplies

* Worksheet with instructions

* Computer paper for showing work for presentation

* Giant sheets of graph paper for graphing presentation

* Markers

* Rulers

latus rectum/

focal chord

latus rectum/ focal chord

Vertex (h, k) Focus

AOS y= k Directrix

Opening – right (p>0) or left (p<0)

Length of the latus rectum

AOS

AOS

Vertex (h, k) Focus

AOS x = h Directrix

Opening – up (p>0) or down (p<0)

Length of the latus rectum

( ),h k p+

y k p= −

( ),h p k+

x h p= −

4p(y – k) = (x - h)2 4p(x – h) = (y - k)2

4 p units

4 p units

NOTE: a = 1/(4p)

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