Chapter 4mm34 Cas

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Chapter 4 – Polynomial Functions

n

n xa xaa x P ...)( 1

10 is a polynomial in x where naaa ,...,, 10 are the

coefficients. If n

n xan ,0 is the leading term and the degree of the

polynomial is n.

Eg. 1253)( 234 x x x x P is a degree 4 polynomial or a quartic function.

We will consider cubic and quartic functions in this chapter.

Firstly we investigate expansions of the expressions of the form nbax .

Summation Notation

n

mi

nmmi nmand J nmwhereaaaa ,,.....1

Expanded form

Also known as sigma notation since ∑ is the upper case form of the Greek

letter sigma.

Eg1. Write

5

1

2

i

i

in expanded form and evaluate.

Eg2. Write 222 30....21 using summation notation.

Ex A2 pg 699 all

Binomial Theorem

The expanded form of nbax can be written in summation notation.

n

k

k k nn

baxk

n

bax0

For example, the first, second and (r+1)th term are,

r r n

r

nnbax

r

nt bax

nt ax

nt

1

1

21 ,1

,0

Eg3. Find the coefficient of 20 x in the expansion of 302 x .



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Eg4. Expand 532 x .

Eg5. Find the 8th term in the expansion of 1042 x .

Ex A3 pg 702 Q1(1/2), Q2(1/2), Q3, Q6(1/2), Q8

Equating Coefficients

If n

n

n

n xb xb xbb xQ xa xa xaa x P ....)(and.....)( 2

210

2

210 are

equal then their corresponding coefficients are equal.

i.e. nn babababa ,....,,,221100

Eg6. If R x for b xa x x 22 346 , find the values of a and b.

Eg7. Show that 866 23 x x x cannot be written in the form bc xa 3

for real numbers a, b and c.

Division of polynomials

This was studied last year. In general for non-zero polynomials,

)()( x Dand x P , if )( x P (the dividend) is divided by )( x D (the divisor),

then there are unique polynomials, )( xQ (the quotient) and )( x R (the

remainder), such that;



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)()()()( x R xQ x D x P

Either the degree of 0)(,)()( x Ror x D x R

When 0)( x R then )( x D is called a divisor of )( x P .

Eg8. Divide 82793 24 x x x by 2 x .

The Remainder Theorem

If a polynomial )( x P is divided by bax the remainder is

a

b P .

The Factor Theorem

bax is a factor of the polynomial )( x P if and only if 0

a

b P .

Eg9. Find the remainder when 123)( 23

x x x x P is divided by 12 x .

Eg10. Given 1 x and 2 x are factors of b xax x x 66 234 , find the

values of a and b.



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Ex 4A: 2ac, 3ac, 4, 6, 8, 11, 12(1/2), 13(1/2), 14, 16(1/2)

Quadratic Functions (Parabolas)General rule cbxax y 2 , to sketch on:

If 0a it has a minimum turning point.

If 0a it has a maximum turning point.

c gives the y intercept.

Axis of symmetry has equationa

b x

2

.

Solve 02 cbxax for x for x intercepts. This may be done by

factorising, completing the square or the quadratic formula,

a

acbb x

2

42

.

Remember - if discriminant 042 acb graph has 2 x intercepts.

- if discriminant 042 acb graph has 1 x intercept.

- if discriminant 042 acb graph has no x intercepts.

Eg11. Sketch 7123)( 2 x x x f showing all intercepts and turning point.

Ex 4B: Q1-4(1/2), 5, 6, 7, 9Ex 4C: Q1(1/2), 3



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k h xa x f 3)(

)1(

)(

2

3

x x

x x x f

)1)(1(

)( 3

x x x

x x x f

y

x

y

x

Cubic Functions of the form

Consider 3)( x x f . It has a zero gradient at (0,0).

We will look at cubics with a dilation and/or

translation applied to 3)( x x f .

Eg12. Sketch 12)( 3

x x f Eg13. Sketch 21)( 3

x x f

Eg14. The graph of k h xa x f 3

)( has zero gradient at (1,1) and

passes through (0,4). Find a, h, k.

Ex 4D: Q1(1/3), 2, 4

Cubics of the form d cxbxax x f 23)(

For now we will look at the various shapes of graphs of cubic functions of

this form. Later on Calculus will be required to determine turning points,

since they do not occur symmetrically between x intercepts as for

quadratics.

Eg15. Eg16.

y

x



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)2()1(

23)(

2

3

x x

x x x f

)2()1(

23)(

2

3

x x

x x x f

y

x

y

x

Eg17. Eg18.

Cubic Equations

You need to be able to solve cubics to find x intercepts. Use remainder

and factor theorems to find factors, then you may need to divide to fullyfactorise.

Eg19. Solve 301140 23 x x x

Ex 4E: Q1-2(1/2), 3

Determining rules for graphs of cubic functions

Eg20. Find the rule for:

Eg21. Find the rule for the cubic function whose graph passes through

(0,1) , (1,4) , (2,17) , (-1,2) of the form d cxbxax x f 23)( .

y

x

4

-3 1 4



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Using TI-Nspire

Or you can set up simultaneous equations first and then solve for a,b, c and d.

Ex 4G: Q1, 2, 3a, 4bd, 5a

Quartic Functions

The general form is edxcxbxax x f 234)( . Here are some graphs of

quartic functions.

y

x

y = x 4

y

x

y = x 4+ x

2

y

x

y = x 4 – x

2

y

x

y = (x – 1)2(x + 2)

2



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Positive quartics will look like or .

Negative quartics will look like or .

Remember if )( x f has factor; 1a x , graph cuts x axis at a x .

2a x , graph bounces off x axis at a x .

3a x , graph has inflection point on x

axis ata x

.Eg22. If 1 x and 2 x are factors of b xax x x x f 66)( 234 , find

a and b.

Ex 4F: 1 (CAS), 2: Ex 4G: Q6: Ex 4D Q3, 5(1/2)

y

x

y = (x – 1)3(x + 2)

y

x

y = (1 – x )(2x + 1)2(4x – 2)

y

x

y = (x + 2)(x + 1)(x – 1)(x – 4)



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Solving Literal Equations (Non-Linear)

Note: If n is an odd natural number, then

i) if nnn aor abab

1

ii) if q p are integers, then

q p p

q

q

p

aor abab

Eg 23. Solve for x

a) a x 7

5

b) ad bx 3

c) 0127 223 xaax x d) 02 k kx x

Simultaneous Equations

Eg 24. Find the coordinates of the points of intersection of each of the

following,

a) b)

Ex 4H Q1 (1/2) 2, 3ac, 4(1/2), 5(1/2), 7, 8, 11, 14a, 15, 17

Review 4 - Odds

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Chapter 4mm34 Cas