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VCE Maths Methods - Unit 1 - Quadratic Functions
Quadratic functions
• Expanding quadratic expressions• Factorising quadratic expressions• Factorisation by inspection• Completing the square• Solving quadratic equations• The quadratic formula• The discriminant (1) • The discriminant (2)• Transformations of parabolas - dilations• Transformations of parabolas - horizontal translations• Transformations of parabolas - vertical translations• Graphing quadratics (1)• Graphing quadratics (2)
VCE Maths Methods - Unit 1 - Quadratic Functions
Expanding quadratic expressions
Each term in one bracket must be multiplied by the terms in the other bracket.
y =(x−1)(x+5) y =2(3x−1)(x−4) y =(2x+5)2
y = x2+5x−x−5
y = x2+4x−5
y =2(3x2 −12x−x+4)
y =2(3x2 −13x+4)
y =6x2 −26x+8
y =4x2+10x+10x+25
y =4x2+20x+25
“FOIL” - first, outside, inside,
last
(ax+b )2 =(a2x2+2abx+b2 )
(ax−b )2 =(a2x2 −2abx+b2 )
y =(x+4)(x−4)
y = x2+4x−4x−16
y = x2 −16
This is a perfectsquare.
This is a difference of two squares.
(ax+b )(ax−b )=(a2x2 −b2 )
VCE Maths Methods - Unit 1 - Quadratic Functions
Factorising quadratic expressions
Find highest common factor
Perfect squares
Factorisation by inspection
(This is now much easier to factorise.)
y = x2+6x+9
This is the square of (x+3)
y =(x+3)2
9 =3 2×3=6
y =60x2+40x+5
y =5(12x2+8x+1) y =5(6x+1)(2x+1)
Difference of two squares
y =49x2 −4
y =72 x2 −22
y =(7x+2)(7x−2)
y =5(12x2+8x+1)
VCE Maths Methods - Unit 1 - Quadratic Functions
Factorisation by inspection
No common factors
y = x2+16x+63
y =16x2 −24x+9
y =6x−24x2
y =72x2 −24x+2
Take out the common factor of 6x
y =6x(1−4x ) y =(x+7)(x+9)
7 & 9 are the only factors of 63
y =16x2 −24x+9Factors of 16: 1 & 16, 2 & 8, 4 & 4
Factors of 9: 1 & 9, 3 & 3
y =(16x− .....)(x− .....) y =(4x− .....)(4x− .....)
y =(4x−1)(4x−9)
y =(4x−3)(4x−3)=(4x−3)2
There are lots of factor pairs from 36, but only one pair from 1.
(And they must both be negative!)
y =(.....x−1)(.....x−1)
y =2(6x−1)2
y =2(36x2 −12x+1)
VCE Maths Methods - Unit 1 - Quadratic Functions
Completing the square
• Any quadratic function can be made into a perfect square form.
• This can be useful to find turning points or to solve difficult equations.
y =2x2+12x+14
y =2(x2+6x+7)
y =2(x2+6x+9−9+7)
y =2 (x+3)2 −2
y =2 (x+3)2 − 2
2
The co-efficient of x2 must be 1, 2 is taken out as a common factor.
The co-efficient of x is halved, squared,added & subtracted. (No change to equation)
y =2 (x2+6x+9)−9+7 The first three terms make a perfect square.
y =2(x+3+ 2)(x+3− 2)
Difference of two squares
Two factors
VCE Maths Methods - Unit 1 - Quadratic Functions
Solving quadratic equations
• The quadratic equation needs to first be factorised.
• Quadratic equations are solved using the Null factor law - if either factor is equal to 0, then the whole equation is equal to 0.
• On the graph, the solutions to the equation y = 0 are the x intercepts.
0=(x−4)2 0=(x−2)(2x+5) 0= x2+6x−12
0= x−2x =2
0=2x+5−5=2x
x =−52
0= x−4x =4
This is a repeated factor & just one solution.
(The graph turns on the x axis, without crossing)
0= x2+6x+9−9−12
0=(x+3)2 −21
0=(x+3)2 −( 21)2
0=(x+3+ 21)(x+3− 21)
0= x+3+ 21
x =−3− 21
0= x+3− 21
x =−3+ 21
This can be a DOTS!
VCE Maths Methods - Unit 1 - Quadratic Functions
The quadratic formula
• The solution to the general quadratic formula (0 = ax2 + bx + c) can be found by completing the square.
• This can be used to find any solutions that exist for a given quadratic.
y = ax2+bx+c
x = −b± b2 −4ac
2a
a=2, b=-4, c=-6 y =2x2 −4x−6For example:
x =
−−4± (−4)2 −(4×2×−6)2×2
x = −b± b2 −4ac
2a
x = 4± 64
4 x = 4±8
4x = -1 & x = 3 (Two solutions)
VCE Maths Methods - Unit 1 - Quadratic Functions
The discriminant
• Quadratic functions can have two, one or no solutions.
• The number of solutions can be determined by the discriminant.
• This is the expression inside the square root in the quadratic equation.
• If, ∆ < 0, there is no solution.
• If, ∆ = 0, there is one solution.
• If, ∆ > 0, there are two solutions.
Δ=b2 −4ac
VCE Maths Methods - Unit 1 - Quadratic Functions
The discriminant (2)
y = x2 −8x+16
Δ=(−8)2 −(4×1×16)=0
y = x2 −6x+11
Δ=(−6)2 −(4×1×11)=−8
y =(x−4)2
y =(x−3)2+2 y =(x−5)2 −4
y = x2 −10x+21
Δ=(−10)2 −(4×1×21)=16
One solution No solutions Two solutions
VCE Maths Methods - Unit 1 - Quadratic Functions
Transformations of parabolas - dilations
y = ax2 : a is the dilation factor that narrows or widens the parabola
(The curve defined by a quadratic function is a parabola.)
VCE Maths Methods - Unit 1 - Quadratic Functions
Transformations of parabolas - horizontal translations
y = (x-h)2 : h is the horizontal translation that moves the graph h units to the right
VCE Maths Methods - Unit 1 - Quadratic Functions
Transformations of parabolas - vertical translations
y = x2 + k : k is the vertical translation that moves the graph up k units
y = x2 y = x2+2
y = x2 −5
VCE Maths Methods - Unit 1 - Quadratic Functions
Graphing quadratics (1)
x intercept: -5 & 1
y =(x−1)(x+5)
x = 1
2(1−5)=−2
y=-3 x 3 = -9
turning point: (-2,-9)
y =(x−1)(x+5)
Turning point:
y =(−2−1)(−2+5)
VCE Maths Methods - Unit 1 - Quadratic Functions
Graphing quadratics (2)
x intercepts:
x intercepts: y = 0
4=(x+1)2
±2= x+1
−1±2= x+1x = -3 and x = 1
turning point: (-1,-4)
y =(x+1)2 −4
y intercept: x = 0, y =-3
0=(x+1)2 −4
VCE Maths Methods - Unit 1 - Quadratic Functions
Finding the equation (1)
1. Decide on turning point or factor form.2.Find the value of a by substituting another
point into the equation.
y = a(x−2)2 −8
Another point (4,0)
Turning point (2,-8)
0= a(4−2)2 −8
8= a(22 )
a =2
y =2(x−2)2 −8
y =2x2 −8x
VCE Maths Methods - Unit 1 - Quadratic Functions
Finding the equation (2)
1. Decide on turning point or factor form.2.Find the value of a by substituting another
point into the equation.
y = a(x−0)(x−4)
y = ax(x−4)x intercepts: 0 & 4
Another point (2,-8)
−8= a(2)(2−4)
−8= a(−4)
a =2
y =2x(x−4)
y =2x2 −8x