3-1 completing the square (presentation)

8/8/2019 3-1 Completing the Square (Presentation)

1/29

Unit 3 Pol nomial Functions

31 Com letin the S uare and S nthetic Division

32 Zeros of Polynomial Functions

33 Polynomial Functions and Their Graphs


2/29

31 Completing the Square

Unit 3 Quadratic and Polynomial Functions


3/29

Conce ts and Ob ectives

Com letin the S uare Ob . #9

Transform a quadratic function into vertex form Review performing synthetic division

Evaluate polynomial functions using the remainder

theorem


4/29

Quadratic Functions

A ol nomial unction of de ree n where n is anonnegative integer, is a function defined by anexpression of the form

( ) = + + + +1

1 1 0...n n

n nf x a x a x a x a

where a b and c are real numbers and a 0.( ) = + +2 f x ax bx c


5/29

Quadratic Functions

As ou ma recall the ra h of a uadratic function is a

parabola. The graph ofg(x) = ax2

is a parabola withvertex at the origin that opens up ifa is positive and

.

determined by the magnitude (absolute value) ofa.

By completingthesquare, any quadratic function can bewritten in the form where (h,k) is the( ) ( )= +

2F x a x h k

.


6/29

Com letin the S uare

Com letin the s uare of a uadratic function allows us

to identify its characteristics and sketch its graph moreeasily.

o comp ete t e square, we a an su tract t e same

number to the function in order to create a perfect

square trinomial, which we can then factor.

We can thus identify the vertex and the axis of symmetry

of the parabola.


7/29


Exam le: Identif the vertex of the arabola

( ) = +2 6 7 f x x x

We want to add a number inside the parentheses that

= + x x x

wou ma e e express on a per ec square. o o s,

we take the square of half the middle term:

2

( ) ( ) = = 6 3 9

2


8/29


Exam le cont

We cant change the balance of the function, so if we

add 9, we also have to subtract 9:

( ) ( )= + +2 6 9 9 7 f x x x ( )= + +2 6 9 9 7x x( )=

23 2x

The vertex is (3, 2). (Notice the signs!)


9/29


Exam le: Gra h b com letin the= +23 2 1 x x x

square and locating the vertex.

To complete the square, the coefficient of 2 must be 1:

( ) = + +2 23 1 f x x x

= =2 2

1 2 1 1

( ) = + + +2

2 1 13 1 f x x x


10/29


Exam le cont. :

Notice that we added the number inside the parentheses.

When we move it outside, we will use the distributive

property:

2 2 1 1= 3 9 9

= 2

1 4

The vertex is at

3 3

1 4,


11/29


12/29


Exam le cont. :

Lets plot the points we have so far:

Plug in values to find other points:

x y

1 4

1

12

3

24


13/29


Exam le cont. :

Finally, we can sketch in the parabola:


14/29

Vertex Formula The vertex form of the uadratic function ives us the

vertex formula:

= =b

Example: Find the vertex of the parabola whose

2a

2

= = 4

12 2

h ( )= =1 3k f

Therefore, the vertex is at(1, 3).


15/29

Dividin Pol nomials Let x and x be ol nomials with x of de ree one

or more, but of lower degree than f(x). There existunique polynomials q(x) and r(x) such that

where either r(x) = 0 or the degree ofr(x) is less than thede ree of x .

= +ix g x q x r x


16/29

Dividin Pol nomials For exam le could be evaluated as

3 23 2 150x x 4x

3 2 + 4 3 2 0 150 x x x x

+ +3 23 0 12 x x x 2

+ 22 0 8x x

12 158x

or +2

12 1583 2

4x

x


17/29

Dividin Pol nomials Usin the division al orithm this means that

( )( ) = + 3 2 23 2 150 4 3 2 12 158 x x x x x

( )f x ( )g x ( )q x ( )r x

=


18/29

S nthetic Division A shortcut method of erformin lon division with

certain polynomials, called syntheticdivision, is usedonly when a polynomial is divided by a binomial of the

, .


19/29

S nthetic Division To see how this works com are the followin :

+ +23 2

3 10 40x x

3 10 40

3 2

2

3 12x x

+

3 12

10 0

2

10 40x x

10 40

40 150

40 160

10

x 40 16010


20/29

S nthetic Division Since the first numbers in each row re eat we can

eliminate them: 1 4003 3 4010

+

123

+

12

4010

40

160

10

40

160

10


21/29

S nthetic Division The entire roblem can now be condensed verticall

and the top row of numbers can be omitted since itduplicates the bottom row if the 3 is brought down:

+

4 3 2 0 150

12

+

4 3 2 0 150

12 40 160

+

10 0

40

3 10 40 10

40 150

160

10


22/29

S nthetic Division It is easier to understand if we chan e the si n of the

divisor (which changes the sign of the numbers in thesecond row) and add instead of subtract:

+ 4 3 2 0 150 + 4 3 2 0 150

3 10 40 10

3 10 40 10

+ + +

2 103 10 404

x xx


23/29

S nthetic Division Exam le: Use s nthetic division to divide

+

3 2

4 15 11 103

x x x x

3 4 15 11 10

4 3 2 4

+ +

2 44 3 23

x x


24/29

S nthetic Division Exam le: Use s nthetic division to divide

+ + +4 3 25 4 3 9 x x x x

3 1 5 4 3 93 6 6 9

1 2 2 3 0

+ +3 22 2 3 x x x


25/29

Remainder Theorem B eneralizin the s nthetic division rocess we can

make the statement that

For an ol nomial x and an com lex

number k, there exists a unique polynomial q(x)

and number rsuch that) ) )= +f x x k q x r

,

( ) ( ) ( ) ( )= + =orf k k k q k r f k r


26/29

Remainder Theorem This roves the followin remainder theorem:

If the polynomialf(x) is divided byx k, then the

Example: Let . Findf(3).

( ) = + 4 2

3 4 5 f x x x x 3 1 0 3 4 5

3 9 1842

1 3 6 1447

= 3 47f


27/29

Potential Zeros Azero of a ol nomial function is a number ksuch that

f(k) = 0. The real number zeros are thexintercepts ofthe graph of the function.

e rema n er t eorem g ves us a qu c way to ec e

a number kis a zero of a polynomial function defined by

f(x). Use synthetic division to findf(k) ; if the remainderis 0, thenf(k) = 0 and kis a zero off(x) .


28/29

Potential Zeros Exam le: Decide whether the iven number kis a zero

off(x): ( ) = + + = 4 3 24 14 36 45; 3 f x x x x x k

3 1 4 14 36 45

3 212145

1 7 7 15 0

, .


29/29

Homework

Colle e Al ebra

Page 315: 1524 (3), 4750, 52 Turn in: 48, 50, 52

Page 326: 550 (5), 54, 55

Turn in: 10, 30, 40, 50, 54

3-1 completing the square (presentation)

Documents