VCE Maths Methods - Unit 1 - Cubic Functions Cubic functions • Expanding cubic expressions • Factorisation by long division • The factor theorem • Graphs of cubic functions
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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
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