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