Cubic functions - Warrnambool College Maths Methods...VCE Maths Methods - Unit 1 - Cubic Functions Expanding a pair of brackets. Expanding cubic expressions Each term in one bracket

VCE Maths Methods - Unit 1 - Cubic Functions

Cubic functions

• Expanding cubic expressions• Factorisation by long division• The factor theorem• Graphs of cubic functions

VCE Maths Methods - Unit 1 - Cubic Functions

Expanding a pair ofbrackets.

Expanding cubic expressions

Each term in one bracket must be multiplied by the terms in the other brackets.For cubic functions, multiply a pair of brackets first.

y =(x!1)(x+5)

y = x(3x!1)(x!4)

y = x2+5x!x!5

y = x2+4x!5

y = x(3x2!12x!x+4)

y = x(3x2!13x+4)

y =3x 3!13x2

+4x“FOIL” - first,

outside, inside, last

y =(x!4)(2x!1)(x+1)

y =2x 3+x2

!x!8x2!4x+4

y =(x!4)(2x2+2x!x!1)

y =(x!4)(2x2+x!1)

y =2x 3!7x2

!5x+4

(a+b )3= a3

+3a2b+3ab2+b 3

Perfect cubes:

VCE Maths Methods - Unit 1 - Cubic Functions

Factorisation by long division

x3 / x = x2x3 - 4x2 + x + 6x - 3

x2

x2 (x-3) = x3 - 3x2x3 - 3x2 + 0x + 0

- x

-x2 + x + 6 -x2 / x = -x-x2 + 3x + 0 -x (x-3) = -x2 + 3x

Subtract to find remainder- 2x + 6- 2x + 6

- 2

0 Remainder = 0

Subtract to find remainder

- 2x / x = -2

• Long division can be performed on polynomials, in a similar way to numbers.• If there zero remainder, then we know that the divisor is a factor of the

polynomial.

- 2(x-3) = -2x + 6Subtract to find remainder

x2 - x - 2 can be further factorised as (x-2)(x+1) x3 - 4x2 + x + 6 = (x-2)(x-3)(x+1)

VCE Maths Methods - Unit 1 - Cubic Functions

The factor theorem

• If (x-a) is a factor of a polynomial p(x), then p(x) / (x-a) will give a remainder of 0.

• Also, p(a) = 0. This is an x intercept on the graph.

• This can be used to find the first factor.

• For the function y = x3 - 4x2 + x + 6:

x y-3 -60-2 -20-1 00 61 42 03 04 10

(x+1) = 0

(x-3) = 0

(x-2) = 0

VCE Maths Methods - Unit 1 - Cubic Functions

Graphs of cubic functions

y = 1

2x(x!3)(x+2)

The steps:1. Use factor theorem to find a factor.2. Divide by the first factor to find the other factors.3. Once factorised, solve y = 0 to find the x intercepts.4. x = 0 to find the y intercepts.

x intercepts: y = 0

12

x =0,x =0

(x!3)=0,x =3

(x+2)=0,x =!2

y intercept: x = 0y = 0

VCE Maths Methods - Unit 1 - Cubic Functions

Graphs of cubic functions

y =!x(x!2)2

x intercept from the factor (x).

y intercept: x = 0

Turning point on the x-axis from repeated factor (x-2)2.

Graph is inverted due to - sign.

VCE Maths Methods - Unit 1 - Cubic Functions

Graphs of cubic functions - inflection points

y = 1

5(x!1)3

+2

Inflection point: (1,2)

x intercept: y = 0

y intercept: x = 0

y =

15

(0!1)3+2

0=

15

(x !1)3+2

y =2!

15

y =

95=1.8

x = !103 +1= !1.15 !10= (x !1)3