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Functions

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Int 2

Functions

Quadratic Functions y = ax2

Quadratics y = ax2 +c

Quadratics y = a(x-b)2

Quadratics y = a(x-b)2 + c

Factorised form y = (x-a)(x-b)

Starter

Int 2

Learning Intention

Success Criteria

- Understand the term function.

- To explain the term function.

- Work out values for a given function.

Int 2

A roll of carpet is 5m wide. It is solid in strips by the area.

If the length of a strip is x m then the area. A square metres,

is given by A = 5x.

The value of A depends on the value of x.

We say A is a function of x. We write :

A(x) =5x

Example

A(1) = 5 x 1 =5

A(2) = 5 x 2 =10

A(t) = 5 x t = 5t

Int 2

Using the formula for the function we can make a table and

draw a graph using A as the y coordinate.

In the case

The graph is a

straight line

We call this a

Linear function.

Int 2

For the following functions write down the gradient

and where the function crosses the y-axis

f(x) = 2x - 1

f(x) = 0.5x + 7

f(x) = -3x

Sketch the following functions.

f(x) = x

f(x) = 2x + 7

f(x) = x +1

Starter

Int 2

Learning Intention

Success Criteria

- To know the properties of a quadratic function.

- To explain the main properties of the basic quadratic function y = ax2
- using graphical methods.

- Understand the links between graphs of the form y = x2 and y = ax2

Int 2

A function of the form

f(x) = a x2 + b x + c

is called a quadratic function

The simplest quadratics have the form

f(x) = a x2

Lets investigate

Quadratic of the form f(x) = ax2

Key Features

Symmetry about x =0

Vertex at (0,0)

The bigger the value of a the steeper the curve.

-x2 flips the curve about x - axis

Int 2

Example

The parabola has the form y = ax2 graph opposite.

The point (3,36) lies on the graph.

Find the equation of the function.

(3,36)

Solution

f(3) = 36

36 = a x 9

a = 36 ÷ 9

a = 4

f(x) = 4x2

Int 2

Q1. Write down the equation of the quadratic.

(2,100)

Solution

f(2) = 100

100 = a x 4

a = 100 ÷ 4

a = 25

f(x) = 25x2

(x-4)(x-3)

Int 2

Learning Intention

Success Criteria

- To know the properties of a quadratic function.
- y = ax2+ c

- To explain the main properties of the basic quadratic function
- y = ax2+ c
- using graphical methods.

- Understand the links between graphs of the form y = x2 and y = ax2 + c

Quadratic of the form f(x) = ax2 + c

Key Features

Symmetry about x = 0

Vertex at (0,C)

a > 0 the vertex (0,C) is a minimum turning point.

a < 0 the vertex (0,C) is a maximum turning point.

Int 2

Example

The parabola has the form y = ax2 + c graph opposite.

The vertex is the point (0,2) so c = 2. The point (3,38)

lies on the graph. Find the equation of the function.

(3,38)

Solution

f(x) = a x2 + c

(0,2)

f(3) = a . 32 + 2

38 = a . 9 +2

a = (38 -2) ÷ 9

f(x) = 4x2 + 2

a = 4

Int 2

Q1. Write down the equation of the quadratic.

(9,81)

Solution

f(9) = 81

81 = a x 9

a = 81 ÷ 9

a = 9

f(x) = 9x2

(x-5)(x-6)

Int 2

Learning Intention

Success Criteria

- To know the properties of a quadratic function.
- y = a(x – b)2

- To explain the main properties of the basic quadratic function
- y = a(x - b)2
- using graphical methods.

- Understand the links between graphs of the form
- y = x2 and y = a(x – b)2

Quadratic of the form f(x) = a(x - b)2

Key Features

Symmetry about x = b

Vertex at (b,0)

Cuts y - axis at x = 0

a > 0 the vertex (b,0) is a minimum turning point.

a < 0 the vertex (b,0) is a maximum turning point.

Int 2

Example

The parabola has the form f(x) = a(x – b)2.

The vertex is the point (2,0) so b = 2. The point (5,36)

lies on the graph. Find the equation of the function.

Solution

f(x) = a (x-b)2

(5,36)

f(5) = a ( 5 -2)2

(2,0)

36 = a × 9

a = 36 ÷ 9

f(x) = 4(x-2)2

a = 4

Int 2

Learning Intention

Success Criteria

- To know the properties of a quadratic function.

- To explain the main properties of the basic quadratic function
- y = a(x-b)2 + c
- using graphical methods.

- Understand the links between the graph of the form
- y = x2
- and
- y = a(x-b)2 + c

Int 2

Every quadratic function can be written in the form

y = a(x-b)2+c

The curve y= f(x) is a parabola

axis of symmetry at x = b

Y - intercept

Vertex or turning point at (b,c)

(b,c)

Cuts y-axis when

x = 0 y = a(x – b)2 + c

x = b

a > 0 minimum turning point

a < 0 maximum turning point

y = a(x-b)2+c

Int 2

Example 1 Sketch the graph y = (x-3)2 + 2

a = 1

b = 3

c = 2

= (3,2)

Vertex / turning point is (b,c)

y

Axis of symmetry at b = 3

(0,11)

y = (0 - 3)2 + 2

= 11

(3,2)

x

y = a(x-b)2+c

Int 2

Example2 Sketch the graph y = -(x+2)2 + 1

a = -1

b = -2

c = 3

= (-2,1)

Vertex / turning point is (b,c)

y

Axis of symmetry at b = -2

(-2,1)

y = -(0 + 2)2 + 1

= -3

x

(0,-3)

y = a(x-b)2+c

Int 2

Example Write down equation of the curve

Given a = 1 or a = -1

a < 0 maximum turning point

a = -1

(-3,5)

Vertex / turning point is (-3,5)

b = -3

(0,-4)

c = 5

y = -(x + 3)2 + 5

Quadratic of the form f(x) = a(x - b)2 + c

Cuts y - axis when x=0

Symmetry about x =b

Vertex / turning point at (b,c)

a > 0 the vertex is a minimum.

a < 0 the vertex is a maximum.

Int 2

Learning Intention

Success Criteria

- To interpret the keyPoints of the factorised form of a quadratic function.

- To show factorised form of a quadratic function.

Int 2

Some quadratic functions can be written in the

factorised form

y = (x - a)(x - b)

The zeros / roots of this function occur when

y = 0 (x - a)(x - b) = 0 x = a and x = b

Note: The a,b in this form are NOT the a,b in the form

f(x) ax2 + bx + c

Q. Find the zeros, axis of symmetry and turning point

for f(x) = (x - 2)(x - 4)

Zero’s at

x = 2 and x = 4

Axis of symmetry

ALWAYS halfway

between

x = 2 and x = 4

x =3

(3,-1)

Y – coordinate - turning point y = (3 - 2)(3 - 4) = -1